$b(t_{R}^{\prime},\beta^{\prime})=\mathbf{w}^{\infty}\prec{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}$,
it follows that

$b(\hat{t}_{R},\hat{\beta})=\Phi_{\mathbf{R}}^{-1}\big{(}b(t_{R}^{\prime},\beta^{\prime})\big{)}\prec\Phi_{\mathbf{R}}^{-1}\big{(}{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}\big{)}={\mathbf{r}}_{i+1}^{-}\mathbb{L}({\mathbf{r}}_{i})^{\infty}.$

Hence, by Proposition 12.3 applied to $\hat{\beta}$ in place of $\beta$,
$\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$ is transitive. We now
define a subshift $\mathcal{K}_{\beta}^{\prime}(t_{R})$ of
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ by

$\mathcal{K}_{\beta}^{\prime}(t_{R}):=\\{\sigma^{n}(\Phi_{{\mathbf{r}}}(\hat{\mathbf{z}})):\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R}),\
n\geq 0\\}.$

As in the proof of Proposition 7.8, $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is
transitive and of full entropy and full Hausdorff dimension in
$\widetilde{\mathcal{K}}_{\beta^{\prime}}(t_{R}^{\prime})=\widetilde{\mathcal{K}}_{\beta}(t_{R})$,
and contains the sequence $b(t_{R},\beta)=b(t_{R}^{\prime},\beta^{\prime})$.
Moreover, 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.

Finally, Proposition 11.6 implies that $h(\widetilde{\mathcal{K}}_{\beta}(t))$
is constant on each interval in $\mathcal{I}$. ∎

##  13\. Gaps between $\beta$-Lyndon intervals

We saw in Subection 5.1 that when $\beta\in\overline{E}$, the $\beta$-Lyndon
intervals are dense in $[0,\tau(\beta)]$. However, Example 5.10 showed that
this need not be the case when $\beta$ lies in the interior of a basic
interval. In this section we show that having gaps between $\beta$-Lyndon
intervals is in fact typical.

Below we fix a basic interval
$[\beta_{\ell},\beta_{*}]=[\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}}]$
generated by a Lyndon word ${\mathbf{S}}\in\Lambda$.

######  Definition 13.1.

We say an open interval $(t_{1},t_{2})$ is a $\beta$-Lyndon gap if

  1. (i)

$(t_{1},t_{2})$ does not intersect any $\beta$-Lyndon intervals; and

  2. (ii)

any open interval $(u_{1},u_{2})$ properly containing $(t_{1},t_{2})$
intersects at least one $\beta$-Lyndon interval.

Note that this definition is indifferent about whether the endpoints $t_{1}$
and $t_{2}$ belong to $\beta$-Lyndon intervals.

We first show that $\beta$-Lyndon gaps always exist when $\beta$ lies in a
basic interval. Most importantly, we show that the entropy
$h(\widetilde{\mathcal{K}}_{\beta}(t))$ stays constant across each
$\beta$-Lyndon gap.

######  Proposition 13.2.

Let $\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$, where
${\mathbf{S}}\in\Lambda$. Then there are infinitely many $\beta$-Lyndon gaps,
and the left endpoint of each of these lies in $\mathscr{E}_{\beta}$.

###### Proof.

Write ${\mathbf{S}}=s_{1}\dots s_{m}$ and $\mathbb{L}({\mathbf{S}})=c_{1}\dots
c_{m}$, and let $j$ be the integer such that
$\mathbb{L}({\mathbf{S}})=s_{j+1}\dots s_{m}s_{1}\dots s_{j}$. Set
$\mathbf{u}:=s_{j+1}\dots s_{m}$. Since $\beta<\beta_{*}$, we have
$\sigma^{m}(\alpha(\beta))\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$,
so there is an integer $N$ such that
$\sigma^{m}(\alpha(\beta))\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{N}0^{\infty}$.
Choose $M\geq N$, and let $t_{M}$ be the point given by

$b(t_{M},\beta)={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{M}\mathbf{u}^{-}\mathbb{L}({\mathbf{S}})^{\infty}.$

This is a valid greedy $\beta$-expansion, because it does not contain the word
$\mathbb{L}({\mathbf{S}})^{+}=\alpha_{1}\dots\alpha_{m}$, where
$(\alpha_{i}):=\alpha(\beta)$. Also, since
$\mathbf{u}^{-}\prec\mathbb{L}({\mathbf{S}})$, it is clear that
$b(t_{M},\beta)\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=b(\tau(\beta),\beta)$,
so $t_{M}<\tau(\beta)$. Furthermore,
$b(t_{M},\beta)\succ\sigma^{m}(\alpha(\beta))$ by the choice of $M$ and $N$.
Finally, for $j\leq i<m$, the properties of Lyndon words yield

$s_{i+1}\dots s_{m}^{-}c_{1}\dots c_{i}\succcurlyeq s_{1}\dots
s_{m-i}c_{1}\dots c_{i}\succ s_{1}\dots s_{m-i}s_{m-i+1}\dots
s_{m}^{-}={\mathbf{s}}^{-},$

so that

$\sigma^{(M+1)m+i-j}(b(t_{M},\beta))=s_{i+1}\dots
s_{m}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\succ b(t_{M},\beta),$

and also,

$\sigma^{(M+1)m-j}(b(t_{M},\beta))=s_{1}\dots
s_{j}\mathbf{u}^{-}\mathbb{L}({\mathbf{S}})^{\infty}={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\succ
b(t_{M},\beta).$

It follows that $\sigma^{n}(b(t_{M},\beta))\succcurlyeq b(t_{M},\beta)$ for
all $n\geq 0$; in other words, $t_{M}\in\mathscr{E}_{\beta}$. Hence, by Lemma
4.5 (and its proof), $t_{M}$ is the limit of an increasing sequence of
endpoints of $\beta$-Lyndon intervals.

On the other hand, if $k>(M+1)|{\mathbf{S}}|+|\mathbf{u}|$, it is easy to see
that $b_{1}\dots b_{k}^{+}\not\in L^{*}(\beta)$ (where we write
$b(t_{M},\beta)=b_{1}b_{2}b_{3}\dots$): If it were, then $b_{1}\dots
b_{k}^{+}$ would have to end in $\mathbb{L}({\mathbf{S}})^{+}$, but
$\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{M}\mathbf{u}^{-}0^{\infty}\succ\alpha(\beta)$
by the choice of $M$. Thus, $t_{M}$ is not a decreasing limit of endpoints of
$\beta$-Lyndon intervals. Finally, $t_{M}$ is not itself a right endpoint of a
$\beta$-Lyndon interval because $b(t_{M},\beta)$ is aperiodic. Hence, $t_{M}$
is a left endpoint of a $\beta$-Lyndon gap.

Since different choices of $M$ give rise to different sequences
$b(t_{M},\beta)$, there are infinitely many $\beta$-Lyndon gaps. ∎

######  Proposition 13.3.

Let $(t_{1},t_{2})$ be a $\beta$-Lyndon gap. Then
$h(\widetilde{\mathcal{K}}_{\beta}(t_{1}))=h(\widetilde{\mathcal{K}}_{\beta}(t_{2}))$.

###### Proof.

By Lemmas 4.5 and 3.1, $(t_{1},t_{2}]\cap\mathscr{E}_{\beta}=\emptyset$. Thus,
if $t\in(t_{1},t_{2})$ then
$\mathcal{K}_{\beta}(t)=\mathcal{K}_{\beta}(t_{2})$, and so
$\widetilde{\mathcal{K}}_{\beta}(t)\backslash\widetilde{\mathcal{K}}_{\beta}(t_{2})$
is at most countable. Hence,
$h(\widetilde{\mathcal{K}}_{\beta}(t))=h(\widetilde{\mathcal{K}}_{\beta}(t_{2}))$.
Letting $t\searrow t_{1}$ and using the continuity of $t\mapsto
h(\widetilde{\mathcal{K}}_{\beta}(t))$, the proposition follows. ∎

######  Corollary 13.4.

If $\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$ for some
${\mathbf{S}}\in\Lambda$, then
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ is infinite.

###### Proof.

This follows directly from Propositions 13.2 and 13.3, since the latter
implies that the left endpoint of a $\beta$-Lyndon gap does not lie in
$\mathscr{B}_{\beta}$. ∎

######  Corollary 13.5.

Let $\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$ for some
${\mathbf{S}}\in\Lambda$, and suppose $\alpha(\beta)$ is eventually periodic.
If the collection $\mathcal{I}$ of non-transitivity windows is empty, then
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ is countably infinite.

###### Proof.

By the assumption and Proposition 11.5,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive for every
$\beta$-Lyndon interval $[t_{L},t_{R}]$. Furthermore, since $\alpha(\beta)$ is
eventually periodic, the subshift $\Sigma_{\beta}$ is sofic and hence
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is sofic as well. Thus, if
$[t_{L},t_{R}]$ and $[t_{L}^{\prime},t_{R}^{\prime}]$ are two $\beta$-Lyndon
intervals with $t_{R}<t_{R}^{\prime}$, then
$h(\widetilde{\mathcal{K}}_{\beta}(t_{R}))>h(\widetilde{\mathcal{K}}_{\beta}(t_{R}^{\prime}))$
by [27, Corollary 4.4.9]. So the only intervals on which
$\dim_{H}K_{\beta}(t)$ is constant are the $\beta$-Lyndon intervals, which
contain no points of $\mathscr{E}_{\beta}$ except the right endpoints, and
$\beta$-Lyndon gaps, which contain no points of $\mathscr{E}_{\beta}$ except
possibly the left endpoints. In other words, each interval on which
$\dim_{H}K_{\beta}(t)$ is constant contains at most one point from
$\mathscr{E}_{\beta}$. Hence,
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ is countable, and by
Corollary 13.4, it is infinite. ∎

Example 10.4 (c) presents a $\beta$ satisfying the hypotheses of Corollary
13.5.

##  14\. Properties of extended $\beta$-Lyndon intervals

Recall the definition of an extended $\beta$-Lyndon interval (EBLI) from
Section 1.

######  Lemma 14.1.

Let $[t_{L}^{*},t_{R}]$ be an EBLI that is not a $\beta$-Lyndon interval. Then
$t_{L}^{*}\in\mathscr{E}_{\beta}$.

###### Proof.

We have to check that

(14.1) $\sigma^{n}(b(t_{L}^{*},\beta))\succcurlyeq
b(t_{L}^{*},\beta)\qquad\forall\,n\geq 0.$

Let $\mathbf{w}$ be the $\beta$-Lyndon word generating the EBLI, so
$b(t_{L}^{*},\beta)=\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}$
and $b(t_{R},\beta)=\mathbf{w}^{\infty}$, where
$m:=\min\\{n:\sigma^{n}(\alpha(\beta))\preccurlyeq\mathbf{w}^{\infty}\\}$.
Write $\mathbf{w}=w_{1}\dots w_{l}$. We first establish that

(14.2) $w_{i+1}\dots
w_{l}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\preccurlyeq(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\qquad\forall\,0\leq
i<l.$

If $l-i\leq m$, then $w_{i+1}\dots
w_{l}^{-}\preccurlyeq\alpha_{1}\dots\alpha_{l-i}^{-}$ since $\mathbf{w}\in
L^{*}(\beta)$, which yields (14.2) for this case.

If $l-i>m$, then we claim that $w_{i+1}\dots
w_{i+m}\preccurlyeq\alpha_{1}\dots\alpha_{m}^{-}$. Suppose to the contrary
that $w_{i+1}\dots w_{i+m}\succcurlyeq\alpha_{1}\dots\alpha_{m}$. We have
$\sigma^{m}(\alpha(\beta))\preccurlyeq\mathbf{w}^{\infty}\preccurlyeq\sigma^{i+m}(\mathbf{w}^{\infty})$
since $\mathbf{w}$ is Lyndon, and hence it follows that

$\alpha(\beta)=\alpha_{1}\dots\alpha_{m}\sigma^{m}(\alpha(\beta))\preccurlyeq
w_{i+1}\dots
w_{i+m}\sigma^{i+m}(\mathbf{w}^{\infty})=\sigma^{i}(\mathbf{w}^{\infty}),$

contradicting that $\mathbf{w}\in L^{*}(\beta)$. Hence, $w_{i+1}\dots
w_{i+m}\preccurlyeq\alpha_{1}\dots\alpha_{m}^{-}$. We can then cancel the
first $m$ digits on both sides of (14.2) and continue inductively until we are
left with a word of length $\leq m$ in front of
$(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}$ in the left hand side, at which
point we know the inequality to be true from the first case above.

For $1\leq j<l$ we have $w_{j+1}\dots w_{l}^{-}\succcurlyeq w_{1}\dots
w_{l-j}$ (since $\mathbf{w}$ is Lyndon), and so (14.2) with $i=l-j$ gives

$\sigma^{j}(b(t_{L}^{*},\beta))=w_{j+1}\dots
w_{l}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\succcurlyeq\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}=b(t_{L}^{*},\beta).$

It remains to check that

(14.3)
$\alpha_{j+1}\dots\alpha_{m}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\succcurlyeq\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\qquad\forall\,0\leq
j<m.$

This is obvious for $j=0$, so assume $1\leq j<m$. If $l\leq m-j$, then
$\alpha_{j+1}\dots\alpha_{j+l}\succcurlyeq\mathbf{w}$ since
$\sigma^{j}(\alpha(\beta))\succ\mathbf{w}^{\infty}$, hence (14.3) holds for
this case.

So suppose $l>m-j$. Then $\sigma^{j}(\alpha(\beta))\succ\mathbf{w}^{\infty}$
implies $w_{1}\dots w_{m-j}\preccurlyeq\alpha_{j+1}\dots\alpha_{m}$. If we had
equality here then by the Lyndon property of $\mathbf{w}$ we would get

$\sigma^{j}(\alpha(\beta))=w_{1}\dots
w_{m-j}\sigma^{m}(\alpha(\beta))\preccurlyeq w_{1}\dots
w_{m-j}\mathbf{w}^{\infty}\preccurlyeq\mathbf{w}^{\infty},$

contradicting the minimality of $m$. Thus, $w_{1}\dots
w_{m-j}\preccurlyeq\alpha_{j+1}\dots\alpha_{m}^{-}$. Together with (14.2)
(taking $i=m-j$), this gives (14.3). ∎

######  Lemma 14.2.

Any two EBLIs are either non-overlapping or else one contains the other.

###### Proof.

Any two $\beta$-Lyndon intervals are disjoint, and a $\beta$-Lyndon interval
$[t_{L},t_{R}]$ and an EBLI $[u_{L}^{*},u_{R}]$ cannot overlap because the
left endpoint $u_{L}^{*}$ lies in $\mathscr{E}_{\beta}$ by Lemma 14.1, and
therefore does not lie in $[t_{L},t_{R}]$.

It remains to consider the case of two EBLIs
$I_{\mathbf{v}}=[\mathbf{v}^{-}(\alpha_{1}\dots\alpha_{m(\mathbf{v})}^{-})^{\infty},\mathbf{v}^{\infty}]$
and
$I_{\mathbf{w}}=[\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-})^{\infty},\mathbf{w}^{\infty}]$;
suppose they overlap without one containing the other. Without loss of
generality,

(14.4)
$\mathbf{v}^{-}\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{v})}^{-}\big{)}^{\infty}\prec\mathbf{w}^{-}\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-}\big{)}^{\infty}\prec\mathbf{v}0^{\infty}\prec\mathbf{v}^{\infty}\prec\mathbf{w}^{\infty},$

where we have the second inequality because
$\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-})^{\infty}$ does not
lie in the $\beta$-Lyndon interval
$[\mathbf{v}0^{\infty},\mathbf{v}^{\infty}]$, as observed at the beginning of
the proof. Since $\mathbf{v}$ and $\mathbf{w}$ are Lyndon,
$\mathbf{v}^{\infty}\prec\mathbf{w}^{\infty}$ implies
$\mathbf{v}^{\infty}\prec\mathbf{w}0^{\infty}$. So $\mathbf{v}^{\infty}$
begins with $\mathbf{w}^{-}$, and hence $m(\mathbf{v})\geq m(\mathbf{w})$. In
fact $\mathbf{v}$ extends $\mathbf{w}^{-}$: It cannot be a prefix of
$\mathbf{w}^{-}$ because
$\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-})^{\infty}\prec\mathbf{v}0^{\infty}$.
Say $\mathbf{v}=\mathbf{w}^{-}u_{1}\dots u_{l}$. Then from (14.4),

(14.5) $u_{1}\dots
u_{l}^{-}\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{v})}^{-}\big{)}^{\infty}\prec\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-}\big{)}^{\infty}\prec
u_{1}\dots u_{l}0^{\infty}.$

Hence the sequence $(\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-})^{\infty}$
begins with $u_{1}\dots u_{l}^{-}$, and the first inequality in (14.5) implies

$\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{v})}^{-}\big{)}^{\infty}\prec\sigma^{l}\big{(}\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-}\big{)}^{\infty}\big{)}.$

But this is impossible, because $m(\mathbf{v})\geq m(\mathbf{w})$ and
$(\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-})^{\infty}=\alpha(\beta^{\prime})$
for some base $\beta^{\prime}$, hence by Lemma 2.1,

$\sigma^{l}\big{(}\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-}\big{)}^{\infty}\big{)}\preccurlyeq\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{w})}^{-}\big{)}^{\infty}\preccurlyeq\big{(}\alpha_{1}\dots\alpha_{m(\mathbf{v})}^{-}\big{)}^{\infty}.$

This contradiction completes the proof. ∎

######  Proposition 14.3.

The union of all EBLIs is dense in $[0,\tau(\beta)]$.

###### Proof.

We know from Proposition 5.9 and Theorems 7.1 and 9.1 that the $\beta$-Lyndon
intervals are dense in $[0,\tau(\beta)]$ when
$\beta\in\overline{E}\cup\bigcup_{{\mathbf{S}}\in\Lambda}\overline{E^{\mathbf{S}}}\cup
E_{\infty}$. So it remains to show that the EBLIs are dense for $\beta$ in the
interior of a basic interval, i.e.
$\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$ with
${\mathbf{S}}\in\Lambda$.

Let $t\in[0,\tau(\beta))$ with greedy $\beta$-expansion
$b(t,\beta)=b_{1}b_{2}\dots$. Let $\mathcal{N}$ denote the set of all indices
$n\in\mathbb{N}$ such that the word $b_{1}\dots b_{n}$ is Lyndon. Since
$b_{1}=0$, the set $\mathcal{N}$ is not empty. There are now three
possibilities:

  1. (1)

$\mathcal{N}$ is finite and $b_{1}\dots b_{n}\in L^{*}(\beta)$ for each
$n\in\mathcal{N}$;

  2. (2)

$\mathcal{N}$ is infinite and $b_{1}\dots b_{n}\in L^{*}(\beta)$ for each
$n\in\mathcal{N}$; or

  3. (3)

$b_{1}\dots b_{n}\not\in L^{*}(\beta)$ for some $n\in\mathcal{N}$.

In the first case, letting $n_{\max}:=\max\mathcal{N}$, we have
$\sigma^{n_{\max}}(b(t,\beta))\preccurlyeq b(t,\beta)$ and so $t$ lies in the
$\beta$-Lyndon interval $[b_{1}\dots b_{n_{\max}},(b_{1}\dots
b_{n_{\max}})^{\infty}]$.

In the second case, $t$ does not lie in any $\beta$-Lyndon interval but is the
limit of left endoints of $\beta$-Lyndon intervals $[b_{1}\dots
b_{n},(b_{1}\dots b_{n})^{\infty}],n\in\mathcal{N}$.

So it remains to deal with the third case. We show that in this case $t$ lies
in some $EBLI$. Let $N$ be the smallest $n\in\mathcal{N}$ for which
$b_{1},\dots b_{n}$ is not $\beta$-Lyndon, and put $\mathbf{w}:=b_{1}\dots
b_{N}$. Note that $\mathbf{w}\not\in L^{*}(\beta)$; we first aim to determine
$\mathbf{w}^{*}$. Observe that, since
$t<\tau(\beta)=\pi_{\beta}({\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty})$,
there is at least one $\beta$-Lyndon word, namely ${\mathbf{S}}$, that is
greater than $\mathbf{w}$; hence $\mathbf{w}^{*}$ is well defined, and we can
write $\mathbf{w}^{*}=b_{1}\dots b_{l}^{+}$ for some $l<N$.

Since $\mathbf{w}\not\in L^{*}(\beta)$, there is an index $k<N$ such that
$b_{k+1}\dots b_{N}b_{1}\dots b_{k}\succcurlyeq\alpha_{1}\dots\alpha_{N}$. Let
$k_{1}$ be the smallest such $k$. In particular $b_{k_{1}+1}\dots
b_{N}\succcurlyeq\alpha_{1}\dots\alpha_{N-k_{1}}$. But, since $b(t,\beta)$ is
a greedy $\beta$-expansion, we also have the reverse inequality. Hence,

(14.6) $b_{k_{1}+1}\dots
b_{N}=\alpha_{1}\dots\alpha_{N-k_{1}},\qquad\mbox{and}\qquad b_{1}\dots
b_{k_{1}}\succcurlyeq\alpha_{N-k_{1}+1}\dots\alpha_{N}.$

Clearly, then, $l\leq k_{1}$ as otherwise $b_{1}\dots b_{l}^{+}$ would include
the forbidden word $\alpha_{1}\dots\alpha_{l-k_{1}}^{+}$.

Suppose $b_{1}\dots b_{k_{1}}^{+}\in L^{*}(\beta)$. Then
$\mathbf{w}^{*}=b_{1}\dots b_{k_{1}}^{+}$, and $m:=\min\\{n\geq
0:\sigma^{n}(\alpha(\beta))\preccurlyeq(\mathbf{w}^{*})^{\infty}\\}\leq
N-k_{1}$, because $\alpha_{N-k_{1}+1}\dots\alpha_{N}\preccurlyeq b_{1}\dots
b_{k_{1}}\prec\mathbf{w}^{*}$. So in this case, we obtain

$\displaystyle(\mathbf{w}^{*})^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}$
$\displaystyle=b_{1}\dots
b_{k_{1}}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\preccurlyeq b_{1}\dots
b_{k_{1}}(\alpha_{1}\dots\alpha_{N-k_{1}}^{-})^{\infty}$
$\displaystyle=b_{1}\dots b_{k_{1}}(b_{k_{1}+1}\dots b_{N}^{-})^{\infty}\prec
b(t,\beta),$

and hence, $t$ lies in the EBLI generated by $\mathbf{w}^{*}$.

If $b_{1}\dots b_{k_{1}}^{+}\in L^{*}(\beta)$, we repeat the process with
$k_{1}$ in place of $N$. Precisely, there is a smallest integer $k_{2}<k_{1}$
such that $b_{k_{2}+1}\dots b_{k_{1}}^{+}b_{1}\dots
b_{k_{2}}\succcurlyeq\alpha_{1}\dots\alpha_{k_{1}}$. We claim that in fact

(14.7) $b_{k_{2}+1}\dots b_{k_{1}}^{+}=\alpha_{1}\dots\alpha_{k_{1}-k_{2}},$

and hence $b_{1}\dots
b_{k_{2}}\succcurlyeq\alpha_{k_{1}-k_{2}+1}\dots\alpha_{k_{1}}$. To see the
claim, suppose by way of contradiction that $b_{k_{2}+1}\dots
b_{k_{1}}\succcurlyeq\alpha_{1}\dots\alpha_{k_{1}-k_{2}}$. Then this must hold
with equality since $b_{k_{2}+1}\dots b_{k_{1}}$ is part of $b(t,\beta)$. By
the Lyndon property of $b_{1}\dots b_{N}$, it then follows that $b_{1}\dots
b_{N}$ ends in $\alpha_{1}\dots\alpha_{k_{1}-k_{2}}$. Thus,
$\alpha_{k_{1}-k_{2}}=b_{N}=1$, whereas $b_{k_{1}}=0$ by the minimality of
$k_{1}$. This contradiction proves the claim.

Suppose we had $k_{1}-k_{2}>N-k_{1}$. Then (14.7) implies $b_{k_{2}+1}\dots
b_{k_{2}+N-k_{1}}=\alpha_{1}\dots\alpha_{N-k_{1}}$, and so by (14.6),

$b_{k_{2}+N-k_{1}+1}\dots
b_{N}\preccurlyeq\alpha_{N-k_{1}+1}\dots\alpha_{N-k_{2}}\preccurlyeq
b_{1}\dots b_{k_{1}-k_{2}},$

contradicting that $b_{1}\dots b_{N}$ is Lyndon. Thus, $k_{1}-k_{2}\leq
N-k_{1}$.

Suppose now that $b_{1}\dots b_{k_{2}}^{+}\in L^{*}(\beta)$. Since
$\alpha_{k_{1}-k_{2}+1}\dots\alpha_{k_{1}}\preccurlyeq b_{1}\dots b_{k_{2}}$,
we have

$\sigma^{k_{1}-k_{2}}(\alpha(\beta))=\alpha_{k_{1}-k_{2}+1}\alpha_{k_{1}-k_{2}+2}\dots\prec(b_{1}\dots
b_{k_{2}}^{+})^{\infty},$

so $m=\min\\{n\geq
0:\sigma^{n}(\alpha(\beta))\preccurlyeq(\mathbf{w}^{*})^{\infty}\\}\leq
k_{1}-k_{2}$. Hence,

$\displaystyle(\mathbf{w}^{*})^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}$
$\displaystyle\preccurlyeq b_{1}\dots
b_{k_{2}}(\alpha_{1}\dots\alpha_{k_{1}-k_{2}}^{-})^{\infty}=b_{1}\dots
b_{k_{1}}(\alpha_{1}\dots\alpha_{k_{1}-k_{2}}^{-})^{\infty}$
$\displaystyle\preccurlyeq b_{1}\dots
b_{k_{1}}(\alpha_{1}\dots\alpha_{N-k_{1}}^{-})^{\infty}\prec b_{1}\dots
b_{k_{1}}\alpha_{1}\dots\alpha_{N-k_{1}}0^{\infty}$ $\displaystyle=b_{1}\dots
b_{N}0^{\infty}\preccurlyeq b(t,\beta),$

where the first equality uses (14.7), and the second inequality uses that
$k_{1}-k_{2}\leq N-k_{1}$. So, $t$ lies in the EBLI generated by
$\mathbf{w}^{*}$.

If $b_{1}\dots b_{k_{2}}^{+}\not\in L^{*}(\beta)$, we repeat the above process
with $k_{2}$ replacing $k_{1}$ and the Lyndon word $b_{1}\dots b_{k_{1}}^{+}$
replacing the Lyndon word $b_{1}\dots b_{N}$. Continuing this way, we obtain a
sequence $N>k_{1}>k_{2}>\dots$. The process must stop after a finite number of
steps, say $p$, and then $\mathbf{w}^{*}=b_{1}\dots b_{k_{p}}^{+}$ and $t$
lies in the EBLI generated by $\mathbf{w}^{*}$. ∎

######  Lemma 14.4.

Assume $\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$ for some
${\mathbf{S}}\in\Lambda$. Then the entropy function $t\mapsto
h(\widetilde{\mathcal{K}}_{\beta}(t))$ is constant in each EBLI that lies
fully in $[0,\tau(\beta)]$.

###### Proof.

The entropy is obviously constant on each $\beta$-Lyndon interval, so consider
an EBLI
$I_{\mathbf{w}}=[\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty},\mathbf{w}^{\infty}]$
inside $[0,\tau(\beta)]$. Since $\beta$-Lyndon intervals are dense in
$[0,\tau(\beta)]$ for
$\beta\in\overline{E}\cup\bigcup_{{\mathbf{S}}\in\Lambda}\overline{E^{\mathbf{S}}}\cup
E_{\infty}$ and EBLIs are non-overlapping by Lemma 14.2, it must be the case
that $\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$ for some
${\mathbf{S}}\in\Lambda$. Moreover, $\mathbf{w}^{\infty}\preccurlyeq
b(\tau(\beta),\beta)={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\prec{\mathbf{S}}^{\infty}$.
So $\mathbf{w}$ and ${\mathbf{S}}$ are two distinct $\beta$-Lyndon words, and
$m=\min\\{n:\sigma^{n}(\alpha(\beta))\leq\mathbf{w}^{\infty}\\}\geq|{\mathbf{S}}|$,
because $\alpha(\beta)$ begins with $\mathbb{L}({\mathbf{S}})^{+}$.

Let $t_{L}^{*}$ and $t_{R}$ be the left and right endpoints of the projection
$\pi_{\beta}(I_{\mathbf{w}})$. It suffices to show that

$h(\widetilde{\mathcal{K}}_{\beta}(t_{L}^{*}))=h(\widetilde{\mathcal{K}}_{\beta}(t_{R})).$

Suppose
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{L}^{*})\backslash\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
Then

(14.8)
$b(t_{L}^{*},\beta)=\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\qquad\forall\,n\geq
0,$

and there is some integer $n_{0}$ such that $\sigma^{n_{0}}(\mathbf{z})\prec
b(t_{R},\beta)=\mathbf{w}^{\infty}$. We claim that $\mathbf{z}$ must end in a
sequence of the form

(14.9)
$\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{p_{1}}\,\alpha_{1}\dots\alpha_{m}\mathbf{w}^{q_{1}}\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{p_{2}}\,\alpha_{1}\dots\alpha_{m}\mathbf{w}^{q_{2}}\dots,$

where $0\leq p_{i},q_{i}\leq\infty$ for all $i$. Since
$\sigma^{n_{0}}(\mathbf{z})\prec\mathbf{w}^{\infty}$, there is some $n_{1}\geq
n_{0}$ such that $\sigma^{n_{1}}(\mathbf{z})$ begins with $\mathbf{w}^{-}$. By
(14.8), it then follows that $\sigma^{n_{1}}(\mathbf{z})$ begins with
$\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{p_{1}}\alpha_{1}\dots\alpha_{m}$
for some $p_{1}\leq\infty$. But now recall that
$\sigma^{m}(\alpha(\beta))\preccurlyeq\mathbf{w}^{\infty}$. Thus, if
$p_{1}<\infty$, then again using (14.8), the word $\alpha_{1}\dots\alpha_{m}$
must be followed by $\mathbf{w}^{q_{1}}\mathbf{w}^{-}$ for some
$q_{1}\leq\infty$. Repeating this argument, we see that $\mathbf{z}$ ends in a
sequence of the form (14.9).

Now let $\mathcal{X}$ denote the set of all sequences of the form (14.9),
together with their shifts. Then $\mathcal{X}$ is a subshift of
$\widetilde{\mathcal{K}}_{\beta}(t_{L}^{*})$. Also, let $\mathcal{Y}$ denote
the set of all infinite concatenations of the words ${\mathbf{S}}$ and
$\mathbf{w}$ and their shifts, that is,

$\mathcal{Y}:=\left\\{\sigma^{n}\big{(}{\mathbf{S}}^{p_{1}}\mathbf{w}^{q_{1}}{\mathbf{S}}^{p_{2}}\mathbf{w}^{q_{2}}\dots\big{)}:\
n\geq 0,\ 0\leq p_{i},q_{i}\leq\infty\ \forall\,i\right\\}.$

If $\mathbf{z}\in\mathcal{Y}$, then
$\sigma^{n}(\mathbf{z})\succcurlyeq\mathbf{w}^{\infty}$ for all $n\geq 0$
since $\mathbf{w}$ and ${\mathbf{S}}$ are both Lyndon and
$\mathbf{w}^{\infty}\prec{\mathbf{S}}^{\infty}$. Furthermore,
$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ for all $n\geq 0$ since
${\mathbf{S}}$ and $\mathbf{w}$ are both $\beta$-Lyndon. Hence, $\mathcal{Y}$
is a subshift of $\widetilde{\mathcal{K}}_{\beta}(t_{R})$. Since neither
$\mathbf{w}$ nor ${\mathbf{S}}$ is a power of the other (else one of them
wouldn’t be Lyndon) and $|{\mathbf{S}}|\leq m$, a comparison of the
definitions of $\mathcal{X}$ and $\mathcal{Y}$ shows that $h(\mathcal{Y})\geq
h(\mathcal{X})$. As a result,

$h(\widetilde{\mathcal{K}}_{\beta}(t_{L}^{*}))\leq\max\\{h(\widetilde{\mathcal{K}}_{\beta}(t_{R})),h(\mathcal{X})\\}\leq\max\\{h(\widetilde{\mathcal{K}}_{\beta}(t_{R})),h(\mathcal{Y})\\}=h(\widetilde{\mathcal{K}}_{\beta}(t_{R})).$

The reverse inequality is obvious. Therefore, the proof is complete. ∎

######  Lemma 14.5.

The closure of every non-transitivity window (that is, every interval in the
collection $\mathcal{I}^{\prime}$) is an EBLI.

###### Proof.

Let $I_{k}=[\underline{t}_{k},\overline{t}_{k})\in\mathcal{I}^{\prime}$ be a
non-transitivity window for
$\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$ with
${\mathbf{S}}\in\Lambda$. Set $m:=j_{k}$. If $\mathbf{v}_{k}\in L^{*}(\beta)$,
then $\mathbf{v}_{k}^{*}=\mathbf{v}_{k}$, and so we have
$b(\underline{t}_{k},\beta)=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}$
and $b(\overline{t}_{k},\beta)=\mathbf{v}_{k}^{\infty}$. By Lemma 11.1,
$\mathbf{v}_{k}^{\infty}\prec{\mathbf{S}}^{\infty}$, and since $\alpha(\beta)$
begins with $\mathbb{L}({\mathbf{S}})^{+}$, this implies that
$\sigma^{n}(\alpha(\beta))\succ\mathbf{v}_{k}^{\infty}$ for all $n<m$.
Therefore, $\overline{I_{k}}$ is an EBLI.

Suppose now that $\mathbf{v}_{k}\not\in L^{*}(\beta)$. Since $I_{k}$ is a non-
transitivity window, $I_{k}\not\subseteq I_{k-1}$ and so
$\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. Thus, by Lemma 10.12, there exist a
nonnegative integer $r$ and a word $\mathbf{u}$ not ending in
$\alpha_{1}\dots\alpha_{j_{k}}^{-}$ such that
$\mathbf{v}_{k}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$,
and

$\mathbf{v}_{k}^{*}=\begin{cases}\mathbf{v}_{k-1}^{*}&\mbox{if $k\geq 2$ and
$\mathbf{u}=\mathbf{v}_{k-1}^{-}$},\\\
\mathbf{u}^{+}&\mbox{otherwise}.\end{cases}$

If $k\geq 2$ and $\mathbf{u}=\mathbf{v}_{k-1}^{-}$, then, since
$j_{k}>j_{k-1}$, it follows that $\mathbf{v}_{k}$ begins with
$\mathbf{v}_{k-1}^{-}\alpha_{1}\dots\alpha_{j_{k-1}}$ and so $I_{k}\subseteq
I_{k-1}$ by Proposition 10.13. This contradicts our assumption that $I_{k}$ is
a non-transitivity window. Hence, this case cannot happen and
$\mathbf{v}_{k}^{*}=\mathbf{u}^{+}$. Now observe that

$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}=\mathbf{u}(\alpha_{1}\dots\alpha_{m}^{-})^{r+1}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}=(\mathbf{v}_{k}^{*})^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}.$

Hence, $\overline{I_{k}}$ is the EBLI generated by $\mathbf{v}_{k}^{*}$. ∎

######  Lemma 14.6.

Let $\mathcal{T}_{R}^{\prime}$ denote the set of all right endpoints of
maximal EBLIs in $[0,\tau(\beta)]$. Then for each
$t\in\mathcal{T}_{R}^{\prime}$, $\widetilde{\mathcal{K}}_{\beta}(t)$ has a
transitive subshift $\mathcal{K}_{\beta}^{\prime}(t)$ of full entropy
containing the sequence $b(t,\beta)$. Moreover, these subshifts can be chosen
so that $\\{\mathcal{K}_{\beta}^{\prime}(t):t\in\mathcal{T}_{R}^{\prime}\\}$
is a strictly descending collection of subshifts, and they are sofic if
$\alpha(\beta)$ is eventually periodic.

###### Proof.

A maximal EBLI either contains a non-transitivity window or else does not
intersect any non-transitivity windows, in view of Lemmas 14.2 and 14.5. Thus,
the result follows from Proposition 11.5 and Theorems 7.1, 9.1 and 12.1. ∎

######  Corollary 14.7.

Assume $\alpha(\beta)$ is eventually periodic. Then the entropy function
$t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$ takes distinct values on
distinct maximal EBLIs in $[0,\tau(\beta)]$.

###### Proof.

Immediate from Lemma 14.6 and [27, Corollary 4.4.9]. ∎

######  Corollary 14.8.

Assume $\alpha(\beta)$ is eventually periodic. Then an EBLI is maximal if and
only if it is not properly contained in the closure of a non-transitivity
window.

###### Proof.

The forward direction is immediate from Lemma 14.5. For the converse, assume
an EBLI $I_{\mathbf{w}}=[t_{L}^{*},t_{R}]$ is not properly contained in the
closure of a non-transitivity window. Suppose there is another EBLI, say
$I_{\mathbf{v}}=[u_{L}^{*},u_{R}]$ that properly contains $I_{\mathbf{w}}$.
Then $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ (resp.
$\widetilde{\mathcal{K}}_{\beta}(u_{R})$) has a transitive sofic subshift
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ (resp.
$\mathcal{K}_{\beta}^{\prime}(u_{R})$) of full entropy that contains the
sequence $b(t_{R},\beta)$ (resp. $b(u_{R},\beta)$), because neither $t_{R}$
nor $u_{R}$ lies in a non-transitivity window. Since
$\mathcal{K}_{\beta}^{\prime}(u_{R})$ is a proper subshift of
$\mathcal{K}_{\beta}^{\prime}(t_{R})$, this implies
$h(\widetilde{\mathcal{K}}_{\beta}(t_{R}))>h(\widetilde{\mathcal{K}}_{\beta}(u_{R}))$,
contradicting that the entropy is constant on $[u_{L}^{*},u_{R}]$. ∎

###### Proof of Theorem 1.4.

Recall the decomposition (2.12). Assume $\alpha(\beta)$ is eventually
periodic. Then $\beta\not\in E$ by Proposition 2.9. Likewise, $\beta\not\in
E^{\mathbf{S}}$ for any ${\mathbf{S}}\in\Lambda$ because if $\beta\in
E^{\mathbf{S}}$, then $\alpha(\beta)=\Phi_{\mathbf{S}}(\alpha(\hat{\beta}))$
for some $\hat{\beta}\in E$, and $\alpha(\hat{\beta})$ is not eventually
periodic, so $\alpha(\beta)$ isn’t either. Furthermore, $\beta\not\in
E_{\infty}$ (see the remark following Proposition 9.2). Thus, by (2.12),
$\beta\in I^{\mathbf{S}}=[\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}]$
for some ${\mathbf{S}}\in\Lambda$.

If $\beta\in\\{\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}\\}$, then the
$\beta$-Lyndon intervals are dense in $[0,\tau(\beta)]$ by Proposition 5.9 and
Theorem 7.1, so the EBLIs in $[0,\tau(\beta)]$ are just the $\beta$-Lyndon
intervals and the conclusion follows from Theorem 8.1.

Finally, if $\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$, the
theorem follows from Proposition 14.3, Lemma 14.4 and Corollary 14.7. ∎

###### Proof of Corollary 1.5.

We first show that ${\mathbf{b}}^{\prime}$ is well defined. First, suppose
$\sigma^{n}({\mathbf{b}})\succ{\mathbf{b}}$ for some $n$, and take $n$ minimal
with this property. Then $b_{n}=0$, and setting
${\mathbf{b}}^{\prime}:=(b_{1}\dots b_{n})^{\infty}$ we see that
$0^{\infty}\prec\sigma^{n}({\mathbf{b}}^{\prime})\preccurlyeq{\mathbf{b}}^{\prime}$
for all $n\geq 0$, and it is easy to see that no sequence between
${\mathbf{b}}^{\prime}$ and ${\mathbf{b}}$ satisfies these inequalities.
Furthermore,
$\Sigma_{\mathbf{a},{\mathbf{b}}}=\Sigma_{\mathbf{a},{\mathbf{b}}^{\prime}}$
for all $\mathbf{a}$. Second, suppose
$\sigma^{n}({\mathbf{b}})\preccurlyeq{\mathbf{b}}$ for all $n\geq 0$, but
${\mathbf{b}}=b_{1}\dots b_{n}0^{\infty}$ with $b_{n}=1$. Setting
${\mathbf{b}}^{\prime}:=(b_{1}\dots b_{n}^{-})^{\infty}$ we see that
$0^{\infty}\prec\sigma^{n}({\mathbf{b}}^{\prime})\preccurlyeq{\mathbf{b}}^{\prime}$
for all $n\geq 0$ by the assumption ${\mathbf{b}}\succ 10^{\infty}$. Again, no
sequence between ${\mathbf{b}}^{\prime}$ and ${\mathbf{b}}$ satisfies these
inequalities. Moreover,
$\Sigma_{\mathbf{a},{\mathbf{b}}^{\prime}}\subseteq\Sigma_{\mathbf{a},{\mathbf{b}}}$
and
$\Sigma_{\mathbf{a},{\mathbf{b}}}\backslash\Sigma_{\mathbf{a},{\mathbf{b}}^{\prime}}$
is at most countable for each $\mathbf{a}$. Thus, in both cases,
${\mathbf{b}}^{\prime}$ is well defined,
$h(\Sigma_{\mathbf{a},{\mathbf{b}}})=h(\Sigma_{\mathbf{a},{\mathbf{b}}^{\prime}})$
for all $\mathbf{a}$, and ${\mathbf{b}}^{\prime}=\alpha(\beta)$ for some
$\beta\in(1,2]$ by Lemma 2.1.

Now assume ${\mathbf{b}}^{\prime}$ is eventually periodic. For simplicity we
relabel ${\mathbf{b}}^{\prime}$ as ${\mathbf{b}}$ in the rest of the proof. If
$[\pi_{\beta}(\mathbf{w}0^{\infty}),\pi_{\beta}(\mathbf{w}^{\infty})]$ is a
plateau of $t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$, then
$[\mathbf{w}^{-}{\mathbf{b}},\mathbf{w}^{\infty}]$ is a plateau of
$\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$ since
$\Sigma_{\mathbf{a},{\mathbf{b}}}=\Sigma_{\mathbf{w}0^{\infty},{\mathbf{b}}}$
for all $\mathbf{a}\in(\mathbf{w}^{-}{\mathbf{b}},\mathbf{w}0^{\infty}]$.
Similarly, if
$[\pi_{\beta}\big{(}\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\big{)},\pi_{\beta}(\mathbf{w}^{\infty})]$
is a plateau of $t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$, then
$[\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty},\mathbf{w}^{\infty}]$
is a plateau of $\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$.

It remains to show that no interval of the form
$[\mathbf{u}^{-}{\mathbf{b}},\mathbf{u}0^{\infty}]$, with $\mathbf{u}$ an
allowed word in $\Sigma_{\beta}$ ending in $1$, is a plateau of
$\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$. (Since
$\pi_{\beta}(\mathbf{u}^{-}{\mathbf{b}})=\pi_{\beta}(\mathbf{u}0^{\infty})$,
the interval $[\mathbf{u}^{-}{\mathbf{b}},\mathbf{u}0^{\infty}]$ collapses to
a point when projecting under $\pi_{\beta}$, so it is at least conceivable
that $[\mathbf{u}^{-}{\mathbf{b}},\mathbf{u}0^{\infty}]$ is a plateau.) This
is equivalent to showing that the point
$t_{0}:=\pi_{\beta}(\mathbf{u}0^{\infty})$ already lies in a plateau of
$t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$, i.e. in an EBLI.

If $\mathbf{u}$ is not Lyndon, then $t_{0}$ lies inside some $\beta$-Lyndon
interval, hence in an EBLI. If $\mathbf{u}\in L^{*}(\beta)$, then $t_{0}$ is
the left endpoint of a $\beta$-Lyndon interval, hence is contained in an EBLI.
Finally, suppose $\mathbf{u}$ is Lyndon but not $\beta$-Lyndon. Then the proof
of Proposition 14.3 shows that $t_{0}$ lies in an EBLI. Hence, in all cases,
the interval $[\mathbf{u}^{-}{\mathbf{b}},\mathbf{u}0^{\infty}]$ already lies
in one of the plateaus $\tilde{I}_{\mathbf{w}}$. ∎

######  Remark 14.9.

The $0$-plateau of $\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$ is
$[b(\tau(\beta),\beta),1^{\infty}]$ as long as $b(\tau(\beta),\beta)$ does not
end in $0^{\infty}$. The only exceptional case is when
$\beta=\beta_{r}^{\mathbf{S}}$ for ${\mathbf{S}}\in\Lambda$, in which case
$b(\tau(\beta),\beta)={\mathbf{S}}0^{\infty}$ and the $0$-plateau of
$\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$ is
$[{\mathbf{S}}^{-}{\mathbf{b}},1^{\infty}]$. (An equivalent result was proved
also by Labarca and Moreira [25].)

##  15\. Proofs of the other main theorems

###### Proof of Theorem 4.4.

Recall the decomposition (2.12). For ${\mathbf{S}}\in F^{*}$, the left
endpoint $\beta_{\ell}^{\mathbf{S}}$ of $I^{{\mathbf{S}}}$ lies in $E_{L}$;
for ${\mathbf{S}}={\mathbf{s}}\bullet{\mathbf{r}}\in\Lambda$ with
${\mathbf{s}}\in F^{*}$ and ${\mathbf{r}}\in\Lambda$,
$\beta_{\ell}^{\mathbf{S}}$ lies in $\overline{E^{\mathbf{s}}}$; and for all
${\mathbf{S}}\in\Lambda$, the right endpoint $\beta_{*}^{\mathbf{S}}$ of
$I^{{\mathbf{S}}}$ lies in $\overline{E^{\mathbf{S}}}$. Thus, the theorem
follows from Theorems 5.1, 6.1, 7.1, 9.1 and 12.1. ∎

###### Proof of Theorem 1.2.

We first show that
$\dim_{H}(\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta})>0$ for almost all
$\beta\in(1,2]$. Fix for the moment a basic interval
$I^{\mathbf{S}}=[\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}]$ where
${\mathbf{S}}\in\Lambda$, and take
$\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$. Then
$\alpha(\beta)$ begins with $\mathbb{L}({\mathbf{S}})^{+}$. Let $k$ be the
integer such that $\sigma^{|{\mathbf{S}}|}(\alpha(\beta))$ begins with
$0^{k}1$. If there is a string of more than $k$ consecutive zeros in
$\alpha(\beta)$ beyond $\mathbb{L}({\mathbf{S}})^{+}0^{k}1$, then it follows
from the construction in Section 10 that the collection $\mathcal{I}$ of non-
transitivity windows is non-empty. It is well known (see [31, p. 678]) that
for Lebesgue-almost all $\beta\in(1,2]$ the sequence $\alpha(\beta)$ contains
arbitrarily long strings of consecutive zeros. Hence for almost all $\beta\in
I^{\mathbf{S}}$ there is at least one non-transitivity window, and for such
$\beta$, $\dim_{H}(\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta})>0$ by
Theorem 11.7. Since the basic intervals $I^{\mathbf{S}}$,
${\mathbf{S}}\in\Lambda$ cover the interval $(1,2]$ up to a set of Lebesgue
measure zero (see [4]), we conclude that
$\dim_{H}(\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta})>0$ for almost all
$\beta\in(1,2]$.

(In fact the above argument can easily be extended to show that for almost all
$\beta\in(1,2]$, there are infinitely many non-transitivity windows.)

Next, we show that for each $k\in\\{0,1,2,\dots\\}\cup\\{\aleph_{0}\\}$ there
are infinitely many $\beta$’s such that
$|\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}|=k$. For finite $k$ this is
an immediate consequence of Theorem 8.1, because for each $k\in\mathbb{N}$ the
set $\Lambda_{k}$ is infinite. For $k=\aleph_{0}$ it follows from Corollary
13.5, since for each ${\mathbf{S}}\in\Lambda$ we can find a base
$\beta\in(\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}})$ such that
$\mathcal{I}=\emptyset$ and $\alpha(\beta)$ is eventually periodic. (For
instance, let $m:=|{\mathbf{S}}|$ and take
$\alpha(\beta)=\mathbb{L}({\mathbf{S}})^{+}0^{m+1}(10)^{\infty}$.)

Finally, we show that, if $\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$
is uncountable, then its Hausdorff dimension is strictly positive. The
intersection of $\mathscr{E}_{\beta}$ with at least one complementary interval
of $\mathscr{B}_{\beta}$, say $[t_{1},t_{2})$, must be uncountable. Since the
intersection of $\mathscr{E}_{\beta}$ with each $\beta$-Lyndon interval or
$\beta$-Lyndon gap contains at most one point, the interval $[t_{1},t_{2})$
must contain infinitely many $\beta$-Lyndon intervals. Let $[t_{L},t_{R}]$ and
$[u_{L},u_{R}]$ be two of them, ordered so that $t_{R}<u_{R}$. Let
$\mathbf{v}$ and $\mathbf{w}$ be the $\beta$-Lyndon words such that
$b(t_{R},\beta)=\mathbf{v}^{\infty}$ and $b(u_{R},\beta)=\mathbf{w}^{\infty}$.
Now consider all sequences of the form

$\mathbf{v}^{3}\mathbf{w}\mathbf{v}^{k_{1}}\mathbf{w}\mathbf{v}^{k_{2}}\mathbf{w}\mathbf{v}^{k_{3}}\mathbf{w}\dots,\qquad
k_{i}\in\\{1,2\\}\ \forall\,i.$

Since $\mathbf{v}^{\infty}\prec\mathbf{w}^{\infty}$, it is easy to see that
each such sequence is the $\beta$-greedy expansion of some point
$t\in(t_{R},u_{R})\cap\mathscr{E}_{\beta}$. Hence, as in the proof of Theorem
11.7, it follows that

$\dim_{H}(\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta})\geq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t_{1},t_{2})\big{)}\geq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap(t_{R},u_{R})\big{)}>0.$

This completes the proof. ∎

###### Proof of Theorem 1.3.

(i) Let $\beta\in(1,2]$ and $t\in\mathscr{B}_{\beta}$. Fix $\varepsilon>0$.
Then by Theorem 1.1,

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

Since $t\in\mathscr{B}_{\beta}$,
$\dim_{H}K_{\beta}(t+\varepsilon)<\dim_{H}K_{\beta}(t)$. Hence, using the
stability of Hausdorff dimension, it follows that
$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,t+\varepsilon)\big{)}=\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}$.
Thus,

$\displaystyle\dim_{H}K_{\beta}(t)$
$\displaystyle=\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}=\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,t+\varepsilon)\big{)}$
$\displaystyle\leq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap(t-\varepsilon,t+\varepsilon)\big{)}\leq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t-\varepsilon,1]\big{)}$
$\displaystyle=\dim_{H}K_{\beta}(t-\varepsilon),$

so letting $\varepsilon\searrow 0$ and using the continuity of
$t\mapsto\dim_{H}K_{\beta}(t)$ gives (i).

(ii) Assume $\alpha(\beta)$ is eventually periodic; then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ (or
$\mathcal{K}_{\beta}^{\prime}(t_{R})$, as appropriate) is sofic for every
$\beta$-Lyndon interval $[t_{L},t_{R}]$. Take $t\in\mathscr{B}_{\beta}$. Since
$\mathscr{B}_{\beta}\subseteq\mathscr{E}_{\beta}$, we have

(15.1)
$\dim_{H}\big{(}\mathscr{B}_{\beta}\cap(t-\varepsilon,t+\varepsilon)\big{)}\leq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap(t-\varepsilon,t+\varepsilon)\big{)}.$

On the other hand, it follows from our results on
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ and the argument in the
proof of Proposition 4.6 that the only intervals with which
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ can have an uncountable
intersection are the non-transitivity windows, if they are present. Since
$\alpha(\beta)$ is eventually periodic, the construction in Section 10 implies
that there are at most finitely many non-transitivity windows. Furthermore,
since $t\in\mathscr{B}_{\beta}$, $t$ cannot be a left endpoint of a non-
transitivity window by Proposition 11.6. Thus, for all sufficiently small
$\varepsilon>0$, the interval $(t,t+\varepsilon)$ does not intersect any non-
transitivity window. Hence it contains at most countably many points of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$, and so

(15.2)
$\dim_{H}\big{(}\mathscr{B}_{\beta}\cap[t,t+\varepsilon)\big{)}\geq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,t+\varepsilon)\big{)}.$

From (15.1), (15.2) and part (i) of the theorem, (ii) follows. ∎

######  Remark 15.1.

The reason why we cannot extend Theorem 1.3 (ii) to all $\beta\in(1,2]$ has to
do with a technical limitation. All our upper bounds for the size of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ depend on [27, Corollary
4.4.9], which states that, if $\mathcal{X}$ is a transitive sofic subshift and
$\mathcal{Y}$ is a proper subshift of $\mathcal{X}$, then
$h(\mathcal{X})>h(\mathcal{Y})$. Without the assumption of $\mathcal{X}$ being
sofic, this conclusion may fail. When $\alpha(\beta)$ is not eventually
periodic, the subshift $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not
necessarily sofic, and therefore the argument in the proof of Proposition 4.6
does not work. We suspect that nonetheless, Theorem 1.3 (ii) holds for all
$\beta\in(1,2]$.

## Acknowledgments

The first author was partially supported by Simons Foundation grant #709869.
The second author was supported by NSFC No. 11971079.

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