$\mathbf{w}\mathbf{v}\mathbf{z}=U_{0}(\tilde{\mathbf{w}}\tilde{\mathbf{v}}\tilde{\mathbf{z}})\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$,
as desired.

Case (B): Suppose ${\mathbf{r}}=U_{1}(\tilde{{\mathbf{r}}})$. Let
$\tilde{\beta}$ be the base such that
$0\alpha(\beta)=U_{1}(0\alpha(\tilde{\beta}))$. Then by Lemma 7.4 we have
$\alpha(\tilde{\beta})=\Phi_{\tilde{{\mathbf{r}}}}(\mathbf{y})$.

Assume without loss of generality that
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$ begins with $0$.
Note that $\alpha(\beta)$ does not contain the word $00$, and $\alpha(\beta)$
does not end in $(01)^{\infty}$, because if it did, then
$\alpha(\tilde{\beta})$ would end in $0^{\infty}$, which is impossible. If
$\mathbf{w}$ begins with $00$, then we can find a positive integer $M$ such
that $(01)^{M}$ does not occur in $\alpha(\beta)$, and then
$\mathbf{w}(01)^{M}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$.

So assume $\mathbf{w}$ begins with $01$. Since $\mathbf{w}$ is Lyndon, we can
then write $\mathbf{w}=U_{1}(\tilde{\mathbf{w}})$, where $\tilde{\mathbf{w}}$
is again Lyndon. (Possibly, $\tilde{\mathbf{w}}=0$.) As before, we find that
$\tilde{\mathbf{w}}$ is in fact $\tilde{\beta}$-Lyndon. Since
$\mathbf{w}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$, we can
deduce in a similar manner as in Case (A) that
$\tilde{\mathbf{w}}\prec\tilde{{\mathbf{r}}}^{-}\mathbb{L}(\tilde{{\mathbf{r}}})^{\infty}$.

Since $\mathbf{z}$ begins with $0$, the same reasoning as before yields that
$\mathbf{z}=U_{1}(\tilde{\mathbf{z}})$ for some sequence $\tilde{\mathbf{z}}$.
As before, we have
$\tilde{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\tilde{\beta}}(\tilde{\mathbf{w}})$.
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 $\mathbf{v}:=U_{1}(\tilde{\mathbf{v}})$; then
$\mathbf{w}\mathbf{v}\mathbf{z}=U_{1}(\tilde{\mathbf{w}}\tilde{\mathbf{v}}\tilde{\mathbf{z}})\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$.
∎

######  Proposition 7.6.

Let $\beta$ be renormalizable, so
$\alpha(\beta)=\Phi_{\mathbf{r}}(\alpha(\hat{\beta}))$ for some Farey word
${\mathbf{r}}$ and base $\hat{\beta}\in(1,2]$. Let
$t_{R}\in\mathcal{T}_{R}(\beta)$. Then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive if and only if
$b(t_{R},\beta)\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.

###### Proof.

Let $\mathbf{w}$ be the $\beta$-Lyndon word such that
$b(t_{R},\beta)=\mathbf{w}^{\infty}$. Set
$\hat{\mathbf{y}}:=\alpha(\hat{\beta})$, so
$\alpha(\beta)=\Phi_{\mathbf{r}}(\hat{\mathbf{y}})$ and $\hat{\mathbf{y}}\succ
10^{\infty}$.

Assume first that
$\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
Then we can find positive integers $N$ and $k$ such that

(7.3)
$\mathbf{w}^{k}1^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{N}0^{\infty}.$

Let $\mathbf{u}=u_{1}\dots
u_{l}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$. By Lemma 7.5 there is a
word $\mathbf{v}$ such that
$\mathbf{w}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$. It
is easy to see that then
$\mathbf{w}^{k}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$
also. Now let $j$ be the smallest positive integer such that $u_{j+1}\dots
u_{l}$ is a prefix of $\mathbf{w}^{\infty}$; or set $j=l$ if no such integer
exists. By choosing $k$ even larger if necessary, we may assume
$k|\mathbf{w}|>l-j$. Consider the sequence

${\mathbf{c}}:=u_{1}\dots u_{j}\mathbf{w}^{k}\mathbf{v}\mathbf{z}.$

Then ${\mathbf{c}}$ begins with $\mathbf{u}$ and ends in $\mathbf{z}$, and
$\mathbf{u}$ and $\mathbf{z}$ are separated at least by the word $\mathbf{v}$.
We now check that ${\mathbf{c}}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.

Since $\alpha(\beta)=\Phi_{\mathbf{r}}(\hat{\mathbf{y}})$, by Lemma 2.12 (iii)
it follows that
$\sigma^{n}(\alpha(\beta))\succcurlyeq\Phi_{\mathbf{r}}(0^{\infty})={\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$
for all $n\geq 0$. Hence, for $n<j$, (7.3) gives

$\sigma^{n}({\mathbf{c}})=u_{n+1}\dots
u_{j}\mathbf{w}^{k}\mathbf{v}\mathbf{z}\prec u_{n+1}\dots
u_{j}{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{N}0^{\infty}\prec\alpha(\beta),$

since $u_{n+1}\dots u_{j}\preccurlyeq\alpha_{1}\dots\alpha_{j-n}$, and
${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{N}0^{\infty}\prec\alpha_{j-n+1}\alpha_{j-n+2}\dots$.
For $n\geq j$, $\sigma^{n}({\mathbf{c}})\prec\alpha(\beta)$ follows since
$\mathbf{w}^{k}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
The other requisite inequality,
$\sigma^{n}({\mathbf{c}})\succcurlyeq\mathbf{w}^{\infty}$, follows from the
choice of $j$. Hence, ${\mathbf{c}}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
This proves that $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive.

Next, assume
$\mathbf{w}^{\infty}\succcurlyeq{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
Note that this actually means
$\mathbf{w}^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$
since $\mathbf{w}^{\infty}$ is periodic. Write
$\alpha(\hat{\beta})=\hat{\alpha}_{1}\hat{\alpha}_{2}\dots$. Since
$\mathbf{w}^{\infty}\prec
b(\tau(\beta),\beta)=\Phi_{\mathbf{S}}(0\hat{\alpha}_{2}\hat{\alpha}_{2}\dots)$,
and the latter expression begins with ${\mathbf{S}}^{-}$ which in turn begins
with ${\mathbf{r}}^{-}$, $\mathbf{w}^{\infty}$ must begin with
${\mathbf{r}}^{-}$. Since $\mathbf{w}$ is also $\beta$-Lyndon and
$\alpha(\beta)=\Phi_{\mathbf{r}}(\hat{\mathbf{y}})$, by Lemma 7.2 it follows
that $\mathbf{w}^{\infty}=\Phi_{\mathbf{r}}(\hat{\mathbf{x}})$ for some
sequence $\hat{\mathbf{x}}$. Again by Lemma 7.2, any sequence
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ that begins with
${\mathbf{r}}^{-}$ must be of the form $\Phi_{\mathbf{r}}(\hat{\mathbf{z}})$
for some sequence $\hat{\mathbf{z}}$. Note that
${\mathbf{r}}^{-}\in\mathcal{L}\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R})\big{)}$
and
$\mathbb{L}({\mathbf{r}})^{\infty}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
But, since each block $\mathbb{L}({\mathbf{r}})$ can only be preceded by
another block $\mathbb{L}({\mathbf{r}})$ or by ${\mathbf{r}}^{-}$, any
sequence of the form
${\mathbf{r}}^{-}\mathbf{v}\mathbb{L}({\mathbf{r}})^{\infty}$ in
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ would have to end in
${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$, which is impossible since
${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}\prec\mathbf{w}^{\infty}$.
This shows $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive. ∎

######  Lemma 7.7.

Let $\beta\in(1,2]$ be a base such that $\alpha(\beta)$ is eventually
periodic, and let $[t_{L},t_{R}]$ be a $\beta$-Lyndon interval. Then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is a sofic subshift.

###### Proof.

It is well known (see [8]) that the $\beta$-shift $\Sigma_{\beta}$ is sofic if
and only if $\alpha(\beta)$ is eventually periodic. Let $\mathbf{w}$ be the
$\beta$-Lyndon word such that $b(t_{R},\beta)=\mathbf{w}^{\infty}$, and define
the set

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

Then $\mathcal{X}_{\mathbf{w}}$ is a SFT and hence sofic. Since the
intersection of two sofic shifts is sofic (see [27, Proposition 3.4.10]), it
follows that
$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\Sigma_{\beta}\cap\mathcal{X}_{\mathbf{w}}$
is sofic. ∎

######  Proposition 7.8.

The statement of Theorem 7.1 (i) holds for all $\beta\in E_{L}^{\mathbf{S}}$,
where ${\mathbf{S}}\in\Lambda$ and
$E_{L}^{\mathbf{S}}:=E^{\mathbf{S}}\cup\bigcup_{{\mathbf{r}}\in
F^{*}}\beta_{\ell}^{{\mathbf{S}}\bullet{\mathbf{r}}}$.

###### Proof.

Write ${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k}$,
where ${\mathbf{r}}_{i}\in F^{*}$ for each $i$. Set
${\mathbf{S}}_{i}:={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{i}$ for
$i=1,\dots,k$, and note that we can write
$\alpha(\beta)=\Phi_{{\mathbf{S}}_{i}}(\alpha(\hat{\beta}_{i}))$ for certain
bases $\hat{\beta}_{1},\dots,\hat{\beta}_{k}$, where in particular
$\hat{\beta}_{k}\in E_{L}$. Let $\mathbf{w}$ be a $\beta$-Lyndon word and
$t_{R}$ the number such that $b(t_{R},\beta)=\mathbf{w}^{\infty}$.

If
$\mathbf{w}^{\infty}\prec{\mathbf{r}}_{1}\mathbb{L}({\mathbf{r}}_{1})^{\infty}={\mathbf{S}}_{1}\mathbb{L}({\mathbf{S}}_{1})^{\infty}$,
then $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive by Proposition
7.6. Otherwise, there is a smallest integer $i\geq 1$ such that
$\mathbf{w}^{\infty}\succ{\mathbf{S}}_{i}\mathbb{L}({\mathbf{S}}_{i})^{\infty}$.
Plainly $i\leq k$. Recall from the proof of Proposition 7.6 that this implies
$\mathbf{w}^{\infty}=\Phi_{{\mathbf{S}}_{i}}(\mathbf{x})$ for some sequence
$\mathbf{x}$. Since $\mathbf{w}^{\infty}$ is periodic and $\mathbf{w}$ begins
with 0 and ends with 1, it follows that the sequence $\mathbf{x}$ is also
periodic, so we can write $\mathbf{x}=\hat{\mathbf{w}}^{\infty}$ for some word
$\hat{\mathbf{w}}$, and it is easy to see that $\hat{\mathbf{w}}$ is
$\hat{\beta}_{i}$-Lyndon, using that $\Phi_{{\mathbf{S}}_{i}}$ is increasing.
Let $\hat{t}_{R}$ be the point such that
$b(\hat{t}_{R},\hat{\beta}_{i})=\hat{\mathbf{w}}^{\infty}$. Now we consider
two cases:

  * •

If $i<k$, then
$\mathbf{w}^{\infty}\prec{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}$,
and so
$\hat{\mathbf{w}}^{\infty}\prec\Phi_{{\mathbf{S}}_{i}}^{-1}\big{(}{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}\big{)}={\mathbf{r}}_{i+1}^{-}\mathbb{L}({\mathbf{r}}_{i+1})^{\infty}$.
Observe that
$\alpha(\hat{\beta}_{i})=\Phi_{{\mathbf{r}}_{i+1}}(\alpha(\hat{\beta}_{i+1}))$.
So by Proposition 7.6,
$\widetilde{\mathcal{K}}_{\hat{\beta}_{i}}(\hat{t}_{R})$ is transitive.

  * •

If $i=k$, then
$\mathbf{w}^{\infty}\prec\tau(\beta)=\Phi_{\mathbf{S}}(0\hat{\alpha}_{1}\hat{\alpha}_{2}\dots)$,
where $(\hat{\alpha}_{i}):=\alpha(\hat{\beta}_{k})$. This implies
$\hat{\mathbf{w}}^{\infty}\prec 0\hat{\alpha}_{1}\hat{\alpha}_{2}\dots$, in
other words, $\hat{t}_{R}<1-(1/\hat{\beta}_{k})$. Observe furthermore that
$\hat{\beta}_{k}\in E_{L}$. Hence,
$\widetilde{\mathcal{K}}_{\hat{\beta}_{k}}(\hat{t}_{R})$ is transitive by
Theorem 5.1.

We conclude that in both cases,
$\widetilde{\mathcal{K}}_{\hat{\beta}_{i}}(\hat{t}_{R})$ is transitive. From
here on we simply write $\hat{\beta}:=\hat{\beta}_{i}$. We also set
$\mathbf{R}:={\mathbf{S}}_{i}$.

Now we set

(7.4)
$\mathcal{K}_{\beta}^{\prime}(t_{R}):=\\{\sigma^{n}(\Phi_{\mathbf{R}}(\hat{\mathbf{z}})):\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R}),\
n\geq 0\\},$

and claim that $\mathcal{K}_{\beta}^{\prime}(t_{R})$ has the desired
properties.

First of all, it is clear that $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is
invariant under $\sigma$, and it is closed because
$\mathcal{K}_{\hat{\beta}}(\hat{t}_{R})$ is closed and
$\Phi_{\mathbf{R}}^{-1}$ is continuous on the range $X(\mathbf{R})$ of
$\Phi_{\mathbf{R}}$. So $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is a subshift of
$\\{0,1\\}^{\mathbb{N}}$.

Next, since each sequence
$\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$
satisfies

$\hat{\mathbf{w}}^{\infty}\preccurlyeq\sigma^{j}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})\qquad\forall
j\geq 0.$

Lemma 2.12 (iii) implies that

$\mathbf{w}^{\infty}=\Phi_{\mathbf{R}}(\hat{\mathbf{w}})^{\infty}\preccurlyeq\sigma^{j}(\Phi_{\mathbf{R}}(\hat{\mathbf{z}}))\preccurlyeq\Phi_{\mathbf{R}}(\alpha(\hat{\beta}))=\alpha(\beta)\qquad\forall
j\geq 0.$

Hence, $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is in fact a subshift of
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$. Furthermore, (7.4) implies that
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ is a factor of
$\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$. If $\alpha(\beta)$ is
eventually periodic, then so is
$\alpha(\hat{\beta})=\Phi_{\mathbf{R}}^{-1}(\alpha(\beta))$; hence
$\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$ is sofic by Lemma 7.7.
But a factor of a sofic subshift is sofic (see [27, Corollary 3.2.2]), and so
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ is sofic.

Note that $b(t_{R},\beta)\in\mathcal{K}_{\beta}^{\prime}(t_{R})$ because
$b(\hat{t}_{R},\hat{\beta})\in\mathcal{K}_{\hat{\beta}}^{\prime}(\hat{t}_{R})$
and $b(t_{R},\beta)=\Phi_{\mathbf{R}}(b(\hat{t}_{R},\hat{\beta}))$.

We next verify that $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is transitive. Let
$\mathbf{u}\in\mathcal{L}(\mathcal{K}_{\beta}^{\prime}(t_{R}))$ and
$\mathbf{z}\in\mathcal{K}_{\beta}^{\prime}(t_{R})$. By the definition of
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ we can extend $\mathbf{u}$ to the left
and right to obtain a word
$\mathbf{u}^{\prime}\in\mathcal{L}(\mathcal{K}_{\beta}^{\prime}(t_{R}))$
containing $\mathbf{u}$ that consists of blocks from
$\\{\mathbf{R}^{-},\mathbf{R},\mathbb{L}(\mathbf{R}),\mathbb{L}(\mathbf{R})^{+}\\}$.
Similarly we can extend $\mathbf{z}$ to the left to a sequence
$\mathbf{z}^{\prime}$ consisting of such blocks. By putting more blocks in
front of $\mathbf{u}^{\prime}$ if necessary, we may assume that
$\mathbf{u}^{\prime}$ begins with the block $\mathbf{R}^{-}$. Also by adding
more blocks to the right of $\mathbf{u}^{\prime}$, we can obtain a word
$\mathbf{u}^{\prime\prime}\in\mathcal{L}(\mathcal{K}_{\beta}^{\prime}(t_{R}))$
ending in $\mathbb{L}(\mathbf{R})^{+}$. Similarly, by putting more blocks in
front of $\mathbf{z}^{\prime}$ we obtain a sequence
$\mathbf{z}^{\prime\prime}\in\mathcal{K}_{\beta}^{\prime}(t_{R})$ beginning
with $\mathbf{R}^{-}\mathbb{L}(\mathbf{R})^{l}$, where $l$ is chosen as large
as possible so that $\mathbf{z}^{\prime\prime}$ is still in
$\mathcal{K}_{\beta}^{\prime}(t_{R})$. (Since
$\mathbf{w}^{\infty}\succ\mathbf{R}^{-}\mathbb{L}(\mathbf{R})^{\infty}$, there
is a largest such $l$.)

Now we will connect the word $\mathbf{u}^{\prime\prime}$ to the sequence
$\mathbf{z}^{\prime\prime}$. Observe that
$\mathbf{u}^{\prime\prime}=\Phi_{\mathbf{R}}(\hat{\mathbf{u}})$ for some word
$\hat{\mathbf{u}}$, and
$\mathbf{z}^{\prime\prime}=\Phi_{\mathbf{R}}(\hat{\mathbf{z}})$ for some
sequence $\hat{\mathbf{z}}$. Since $\Phi_{\mathbf{R}}$ is strictly increasing,
$\hat{\mathbf{u}}\in\mathcal{L}\big{(}\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})\big{)}$
and $\hat{\mathbf{z}}\in\mathcal{K}_{\hat{\beta}}^{\prime}(\hat{t}_{R})$.
Moreover, $\hat{\mathbf{u}}$ begins with $0$ and ends in $1$. Similarly,
$\hat{\mathbf{z}}$ begins with $0^{l+1}$ and extending $\hat{\mathbf{z}}$ by
another 0 would create a sequence that is not in
$\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$.

Since $\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$ is transitive,
there is a (possibly empty) word $\hat{\mathbf{v}}$ such that
$\hat{\mathbf{u}}\hat{\mathbf{v}}\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$.
By the above remark, $\hat{\mathbf{v}}$ must end with $1$. We can write
$\hat{\mathbf{v}}=1^{k}\hat{\mathbf{v}}^{\prime}$, where $k\geq 0$ and
$\hat{\mathbf{v}}^{\prime}$ is either empty or begins with $0$. Now the
sequence $(\hat{\mathbf{u}}1^{k},\hat{\mathbf{v}}^{\prime},\hat{\mathbf{z}})$
is connectible, and hence (see Lemma 2.13)

$\Phi_{\mathbf{R}}\big{(}(\hat{\mathbf{u}}1^{k})\hat{\mathbf{v}}^{\prime}\hat{\mathbf{z}}\big{)}=\Phi_{\mathbf{R}}(\hat{\mathbf{u}}1^{k})\Phi_{\mathbf{R}}(\hat{\mathbf{v}}^{\prime})\Phi_{\mathbf{R}}(\hat{\mathbf{z}}).$

Set $\mathbf{v}:=\mathbf{R}^{k}\Phi_{\mathbf{R}}(\hat{\mathbf{v}}^{\prime})$.
Then
$\mathbf{u}^{\prime\prime}\mathbf{v}\mathbf{z}^{\prime\prime}=\Phi_{\mathbf{R}}(\hat{\mathbf{u}}1^{k}\hat{\mathbf{v}}^{\prime}\hat{\mathbf{z}})=\Phi_{\mathbf{R}}(\hat{\mathbf{u}}\hat{\mathbf{v}}\hat{\mathbf{z}})\in\mathcal{K}_{\beta}^{\prime}(t_{R})$.
This shows that $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is transitive.

Next, we verify that $\mathcal{K}_{\beta}^{\prime}(t_{R})$ has full entropy
and full Hausdorff dimension in $\widetilde{\mathcal{K}}_{\beta}(t_{R})$. If
$\mathbf{z}=z_{1}z_{2}z_{3}\dots\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$,
then

$\mathbf{R}^{-}0^{\infty}\prec\Phi_{\mathbf{R}}(\hat{\mathbf{w}}^{\infty})=\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)=\Phi_{\mathbf{R}}(\alpha(\hat{\beta}))\prec\mathbb{L}(\mathbf{R})^{+}1^{\infty}\qquad\forall
n\geq 0.$

So either $\mathbf{z}\in\Gamma(\mathbf{R})$, where $\Gamma(\mathbf{R})$ is the
set defined in Lemma 7.3, or else $\mathbf{z}$ contains the word
$\mathbf{R}^{-}$ or $\mathbb{L}(\mathbf{R})^{+}$ somewhere, in which case
there is by Lemma 7.2 an $n\in\mathbb{N}$ such that
$\sigma^{n}(\mathbf{z})=\Phi_{\mathbf{R}}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$, and
the prefix $z_{1}\dots z_{n}$ is a word in $\mathcal{L}(\Gamma(\mathbf{R}))$.
This implies

$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R})\big{)}=\max\left\\{h(\Gamma(\mathbf{R})),h\big{(}\mathcal{K}_{\beta}^{\prime}(t_{R})\big{)}\right\\}.$

But $\Gamma(\mathbf{R})$ is countable by Lemma 7.3. Hence, we obtain

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

As a result, $\mathcal{K}_{\beta}^{\prime}(t_{R})$ has full entropy. By the
same reasoning, we also have that

$\dim_{H}\mathcal{K}_{\beta}^{\prime}(t_{R})=\dim_{H}\widetilde{\mathcal{K}}_{\beta}(t_{R}).$

Finally, it is easy to see from (7.4) that
$\\{\mathcal{K}_{\beta}^{\prime}(t_{R}):t_{R}\in\mathcal{T}_{R}(\beta)\\}$ is
a strictly descending collection of subshifts because
$\\{\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R}):\hat{t}_{R}\in\mathcal{T}_{R}(\hat{\beta})\\}$
is one. ∎

######  Proposition 7.9.

The statement of Theorem 7.1 (i) holds for $\beta=\beta_{*}^{\mathbf{S}}$, for
all ${\mathbf{S}}\in\Lambda$.

###### Proof.

The proof is the same as the previous one, with a minor modification. Here we
may assume that
${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k}$ with $k\geq
2$ and each ${\mathbf{r}}_{i}\in F^{*}$, since for ${\mathbf{S}}\in F^{*}$ we
already know $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive for all
$t_{R}\in\mathcal{T}_{R}(\beta)$ by Theorem 6.1. Set
${\mathbf{S}}_{i}:={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{i}$ for
$i=1,\dots,k$, and write
$\alpha(\beta)=\Phi_{{\mathbf{S}}_{i}}(\alpha(\hat{\beta}_{i}))$ for
$i=1,\dots,k-1$, and certain bases $\hat{\beta}_{1},\dots,\hat{\beta}_{k-1}$.
Observe, however, that
$\alpha(\beta)=\mathbb{L}({\mathbf{S}}_{k})^{+}{\mathbf{S}}_{k}^{-}\mathbb{L}({\mathbf{S}}_{k})^{\infty}=\Phi_{{\mathbf{S}}_{k}}(10^{\infty})$
so there is no corresponding base $\hat{\beta}_{k}$.

Let $\mathbf{w}$ be a $\beta$-Lyndon word and $t_{R}$ the corresponding point
such that $b(t_{R},\beta)=\mathbf{w}^{\infty}$. If
$\mathbf{w}^{\infty}\prec{\mathbf{r}}_{1}\mathbb{L}({\mathbf{r}}_{1})^{\infty}={\mathbf{S}}_{1}\mathbb{L}({\mathbf{S}}_{1})^{\infty}$,
then $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive by Proposition
7.6. Otherwise, there is a unique integer $i\leq k-1$ such that

(7.5)
${\mathbf{S}}_{i}^{-}\mathbb{L}({\mathbf{S}}_{i})^{\infty}\prec\mathbf{w}^{\infty}\prec{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty},$

because
$\tau(\beta)=\pi_{\beta}\big{(}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\big{)}$.
Let $\hat{t}_{R}$ be the point such that
$b(\hat{t}_{R},\hat{\beta}_{i})=\hat{\mathbf{w}}^{\infty}$.

  * •

If $i<k-1$, then
$\hat{\mathbf{w}}^{\infty}\prec{\mathbf{r}}_{i+1}^{-}\mathbb{L}({\mathbf{r}}_{i+1})^{\infty}$
by (7.5). Observe that
$\alpha(\hat{\beta}_{i})=\Phi_{{\mathbf{r}}_{i+1}}(\alpha(\hat{\beta}_{i+1}))$.
So by Proposition 7.6,
$\widetilde{\mathcal{K}}_{\hat{\beta}_{i}}(\hat{t}_{R})$ is transitive.

  * •

If $i=k-1$, then
$\hat{\mathbf{w}}^{\infty}\prec{\mathbf{r}}_{k}^{-}\mathbb{L}({\mathbf{r}}_{k})^{\infty}$.
Also $\beta_{k-1}=\beta_{*}^{{\mathbf{r}}_{k}}$. Hence,
$\widetilde{\mathcal{K}}_{\hat{\beta}_{k-1}}(\hat{t}_{R})$ is transitive by
Theorem 6.1.

In both cases, $\widetilde{\mathcal{K}}_{\hat{\beta}_{i}}(\hat{t}_{R})$ is
transitive. The rest of the proof is now the same as before. ∎

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

In order to complete the proof of Theorem 7.1, we need the following extension
of Proposition 5.9.

######  Proposition 7.10.

Let ${\mathbf{S}}\in\Lambda$, and $\beta\in\overline{E^{\mathbf{S}}}$. Then
the $\beta$-Lyndon intervals are dense in $[0,\tau(\beta)]$.

The proof uses the following lemma.

######  Lemma 7.11.

Let ${\mathbf{S}}\in\Lambda$, and let bases $\beta$ and $\hat{\beta}$ be
related by $\alpha(\beta)=\Phi_{\mathbf{S}}(\alpha(\hat{\beta}))$. Define the
map

(7.6)
$\Theta_{{\mathbf{S}},\beta}:[0,1)\to[0,1);\qquad\hat{t}\mapsto\pi_{\beta}\circ\Phi_{\mathbf{S}}\circ
b(\hat{t},\hat{\beta}).$

If $[\hat{t}_{L},\hat{t}_{R}]$ is a $\hat{\beta}$-Lyndon interval, then
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})]$
is a $\beta$-Lyndon interval containing the point
$\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L})$.

###### Proof.

Let $\hat{\mathbf{w}}$ be the $\hat{\beta}$-Lyndon word such that
$b(\hat{t}_{L},\hat{\beta})=\hat{\mathbf{w}}0^{\infty}$. Write
$\alpha(\hat{\beta})=\alpha_{1}\alpha_{2}\alpha_{3}\dots$. Consider the
sequence $(\hat{\mathbf{z}}_{n})$ defined by

$\hat{\mathbf{z}}_{n}:=\hat{\mathbf{w}}^{-}\alpha_{1}\alpha_{2}\dots\alpha_{n}0^{\infty},\qquad
n\in\mathbb{N},$

and set $t_{n}:=\pi_{\hat{\beta}}(\hat{\mathbf{z}}_{n})$. It is clear that
$t_{n}\nearrow\pi_{\hat{\beta}}(\hat{\mathbf{w}}^{-}\alpha(\hat{\beta}))=\pi_{\hat{\beta}}(\hat{\mathbf{w}}0^{\infty})=\hat{t}_{L}$,
and that $\hat{\mathbf{z}}_{n}$ is the greedy $\hat{\beta}$-expansion of
$t_{n}$. So, using that $\Theta_{{\mathbf{S}},\beta}$ is strictly increasing
and the maps $\pi_{\beta}$ and $\Phi_{\mathbf{S}}$ are continuous, we obtain

$\displaystyle\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-)$
$\displaystyle=\lim_{t\nearrow\hat{t}_{L}}\Theta_{{\mathbf{S}},\beta}(t)=\lim_{n\to\infty}\pi_{\beta}\circ\Phi_{\mathbf{S}}(\hat{\mathbf{z}}_{n})$
$\displaystyle=\pi_{\beta}\circ\Phi_{\mathbf{S}}(\hat{\mathbf{w}}^{-}\alpha(\hat{\beta}))=\pi_{\beta}\big{(}\Phi_{\mathbf{S}}(\hat{\mathbf{w}}^{-})\Phi_{\mathbf{S}}(\alpha(\hat{\beta}))\big{)}$
$\displaystyle=\pi_{\beta}\big{(}\Phi_{\mathbf{S}}(\hat{\mathbf{w}})^{-}\alpha(\beta)\big{)}=\pi_{\beta}\big{(}\Phi_{\mathbf{S}}(\hat{\mathbf{w}})0^{\infty}\big{)}.$

Moreover,
$\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})=\pi_{\beta}\big{(}\Phi_{\mathbf{S}}(\hat{\mathbf{w}}^{\infty})\big{)}$.
Thus,
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})]$
is the $\beta$-Lyndon interval
$[\pi_{\beta}(\mathbf{w}0^{\infty}),\pi_{\beta}(\mathbf{w}^{\infty})]$, where
$\mathbf{w}:=\Phi_{\mathbf{S}}(\hat{\mathbf{w}})$. That this interval contains
the point $\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L})$ follows since
$\Phi_{\mathbf{S}}(\hat{\mathbf{w}})0^{\infty}\prec\Phi_{\mathbf{S}}(\hat{\mathbf{w}}){\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=\Phi_{\mathbf{S}}(\hat{\mathbf{w}}0^{\infty})\preccurlyeq\Phi_{\mathbf{S}}(\hat{\mathbf{w}})^{\infty}.$
∎

###### Proof of Proposition 7.10.

We set
$t_{*}:=\pi_{\beta}\big{(}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\big{)}$,
and divide the proof in two parts.

Step 1. The interval $[t_{*},\tau(\beta)]$. If $\beta=\beta_{*}^{\mathbf{S}}$,
then $\tau(\beta)=t_{*}$ so this interval degenerates to a point and there is
nothing to prove. So assume
$\beta\in\overline{E^{\mathbf{S}}}\backslash\\{\beta_{*}^{\mathbf{S}}\\}$.
Then $\alpha(\beta)=\Phi_{\mathbf{S}}(\alpha(\hat{\beta}))$ for some
$\hat{\beta}\in\overline{E}$. Using the map $\Theta_{{\mathbf{S}},\beta}$ from
Lemma 7.11, we observe that $t_{*}=\Theta_{{\mathbf{S}},\beta}(0)$ and
$\tau(\beta)=\Theta_{{\mathbf{S}},\beta}(\tau(\hat{\beta}))$ (see [4,
Proposition 6.2]). Proposition 5.9 tells us that the $\hat{\beta}$-Lyndon
intervals are dense in $[0,\tau(\hat{\beta})]$. Since the Lyndon intervals
have the property that any two of them are either disjoint or one contains the
other (see [22, Proposition 4.1]), it follows that every Lyndon interval in
$[0,\tau(\hat{\beta})]$ is contained in a $\hat{\beta}$-Lyndon interval.

Although the function $\Theta_{{\mathbf{S}},\beta}$ is not continuous (it
jumps where the greedy expansion $t\mapsto b(t,\hat{\beta})$ does), we claim
that the intervals
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})]$,
as $[\hat{t}_{L},\hat{t}_{R}]$ ranges over the $\hat{\beta}$-Lyndon intervals,
cover all the jumps of $\Theta_{{\mathbf{S}},\beta}$. Note that
$\Theta_{{\mathbf{S}},\beta}$ jumps only at points of the form
$\pi_{\hat{\beta}}(\mathbf{w}0^{\infty})$, where $\mathbf{w}$ is a word ending
in 1 such that $\mathbf{w}0^{\infty}$ is a greedy $\hat{\beta}$-expansion. Let
${\mathbf{r}}$ be the longest prefix of $\mathbf{w}$ that is Lyndon. Then the
Lyndon interval
$[\pi_{\hat{\beta}}({\mathbf{r}}0^{\infty}),\pi_{\hat{\beta}}({\mathbf{r}}^{\infty})]$
contains $\pi_{\hat{\beta}}(\mathbf{w}0^{\infty})$. Since this Lyndon interval
is contained in some $\hat{\beta}$-Lyndon interval
$[\hat{t}_{L},\hat{t}_{R}]$, and
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})]$
contains the “jump”
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L})]$
by Lemma 7.11, the claim follows. But then the intervals
$[\Theta_{{\mathbf{S}},\beta}(\hat{t}_{L}-),\Theta_{{\mathbf{S}},\beta}(\hat{t}_{R})]$,
which are $\beta$-Lyndon by Lemma 7.11, are dense in $[t_{*},\tau(\beta)]$.

Step 2. The interval $[0,t_{*})$. Write
$\alpha(\beta)=\alpha_{1}\alpha_{2}\alpha_{3}\dots$, and
${\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=\gamma_{1}\gamma_{2}\gamma_{3}\dots$.
Let $t\in(0,t_{*})$, so $(t_{i}):=b(t,\beta)\prec(\gamma_{i})$. We will show
that

$\sigma^{n}((t_{1}\dots t_{k})^{\infty})\prec\alpha(\beta)\qquad\forall\,n\geq
0,\quad\forall\,k\in\mathbb{N}.$

The density of the $\beta$-Lyndon intervals then follows in the same way as in
the proof of Proposition 5.9.

Fix $k\in\mathbb{N}$ and $0\leq n<k$. Since $(t_{i})$ is a greedy
$\beta$-expansion, $t_{n+1}\dots
t_{k}\preccurlyeq\alpha_{1}\dots\alpha_{k-n}$. We will now show that
$(t_{1}\dots t_{k})^{\infty}\prec\sigma^{k-n}(\alpha(\beta))$. Since
$\beta\in\overline{E^{\mathbf{S}}}$, it follows that $\alpha(\beta)\in
X({\mathbf{S}})$ and so
$\sigma^{l}(\alpha(\beta))\succcurlyeq{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=\gamma_{1}\gamma_{2}\ldots$
for all $l\geq 0$. Thus, it suffices to show that

(7.7) $(t_{1}\dots
t_{k})^{\infty}\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}.$

Let $m:=|{\mathbf{S}}|$. Observe that $t_{1}\dots
t_{k}0^{\infty}\preccurlyeq(t_{i})\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$.
If $k\geq m$, then we see that $t_{1}\dots
t_{m}\preccurlyeq{\mathbf{S}}^{-}\prec\gamma_{k+1}\dots\gamma_{k+m}$, giving
(7.7).

So suppose $k<m$. Let $q$ and $r$ be integers such that $m=qk+r$ and $0\leq
r<k$. Then since ${\mathbf{S}}$ is Lyndon,

$t_{1}\dots t_{k}\preccurlyeq s_{1}\dots s_{k}\preccurlyeq s_{pk+1}\dots
s_{(p+1)k},\qquad p=0,1,\dots,q-1,$

and

$t_{1}\dots t_{r}\preccurlyeq s_{1}\dots s_{r}\preccurlyeq s_{qk+1}\dots
s_{m}^{-}.$

Thus,

$(t_{1}\dots t_{k})^{q}t_{1}\dots t_{r}\preccurlyeq s_{1}\dots s_{m}^{-},$

and if this holds with equality, then

$t_{r+1}\dots t_{k}(t_{1}\dots t_{k})^{q-1}t_{1}\dots t_{r}=s_{r+1}\dots
s_{m}^{-}\prec s_{r+1}\dots
s_{m}\preccurlyeq\mathbb{L}({\mathbf{S}})_{1:m-r},$

where the last inequality follows from the definition of
$\mathbb{L}({\mathbf{S}})$. Putting these results together, we obtain

$(t_{1}\dots t_{k})^{2q}t_{1}\dots
t_{r}\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})_{1:m-r},$

yielding (7.7). ∎

###### Proof of Theorem 7.1.

Statement (i) follows Propositions 7.8 and 7.9, and (ii) follows from
Proposition 7.10. ∎

##  8\. Comparison of the bifurcation sets

Our aim in this section is to prove the following extension of Corollary 5.11.

######  Theorem 8.1.

Let
${\mathbf{S}}={\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2}\bullet\dots\bullet{\mathbf{r}}_{k}\in\Lambda$,
where ${\mathbf{r}}_{i}\in F^{*}$ for each $i$. Suppose
$\beta\in\\{\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}},\beta_{r}^{\mathbf{S}}\\}$.

  1. (i)

If $\beta=\beta_{\ell}^{\mathbf{S}}$, then
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ consists exactly of the
$k-1$ points whose greedy $\beta$-expansions, in decreasing order, are

(8.1)
${\mathbf{r}}_{1}^{\infty},\qquad({\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2})^{\infty},\qquad\dots,\qquad({\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2}\bullet\dots\bullet{\mathbf{r}}_{k-1})^{\infty}.$

  2. (ii)

If $\beta=\beta_{*}^{\mathbf{S}}$ or $\beta_{r}^{\mathbf{S}}$, then
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ consists exactly of the $k$
points whose greedy $\beta$-expansions, in decreasing order, are

${\mathbf{r}}_{1}^{\infty},\qquad({\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2})^{\infty},\qquad\dots,\qquad({\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2}\bullet\dots\bullet{\mathbf{r}}_{k})^{\infty}.$

The proof uses the following lemma. Recall the definition of
$\Gamma({\mathbf{r}})$ from (7.2).

######  Lemma 8.2.

Let ${\mathbf{r}},{\mathbf{s}}\in\Lambda$, and suppose
$\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}$ satisfies

(8.2)
${\mathbf{r}}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{r}}\bullet{\mathbf{s}})^{+}({\mathbf{r}}\bullet{\mathbf{s}})^{\infty}\qquad\forall
n\geq 0.$

Then $\mathbf{z}\in\Gamma({\mathbf{r}})$.

###### Proof.

Since $\mathbb{L}({\mathbf{r}}\bullet{\mathbf{s}})^{+}$ begins with
$\mathbb{L}({\mathbf{r}})^{+}$ and ${\mathbf{r}}\bullet{\mathbf{s}}$ begins
with ${\mathbf{r}}^{-}$, the word
$\mathbb{L}({\mathbf{r}}\bullet{\mathbf{s}})^{+}({\mathbf{r}}\bullet{\mathbf{s}})^{\infty}$
begins with $\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{k}{\mathbf{r}}^{-}$ for
some finite $k$. Hence,
$\sigma^{n}(\mathbf{z})\prec\mathbb{L}({\mathbf{r}})^{+}1^{\infty}$ for all
$n\geq 0$, and if $\mathbf{z}$ ever contains $\mathbb{L}({\mathbf{r}})^{+}$,
then the next block of length $(k+1)|{\mathbf{r}}|$ must be
$\preccurlyeq{\mathbf{r}}^{k}{\mathbf{r}}^{-}$ and also
$\succcurlyeq{\mathbf{r}}^{k+1}$ by the lower bound in (8.2). This is
impossible, so $\mathbf{z}\in\Gamma({\mathbf{r}})$. ∎

###### Proof of Theorem 8.1.

We begin with (i). For $k=1$ this is simply Corollary 5.11, so assume $k\geq
2$. Let
${\mathbf{S}}={\mathbf{r}}_{1}\bullet{\mathbf{r}}_{2}\bullet\cdots\bullet{\mathbf{r}}_{k}\in\Lambda$,
where ${\mathbf{r}}_{1},\dots,{\mathbf{r}}_{k}\in F^{*}$. Let
$\beta=\beta_{\ell}^{\mathbf{S}}$. From Theorem 7.1 and Proposition 4.6 it
follows that
$\mathscr{E}_{\beta}\cap[0,\tau(\beta))=\mathscr{B}_{\beta}\cap[0,\tau(\beta))$,
so it suffices to show that $\mathscr{E}_{\beta}$ has precisely $k-1$ points
in the interval $[\tau(\beta),1)$, whose greedy $\beta$-expansions are given
by (8.1).

Observe from [4, Theorem 2] that
$\tau(\beta)=\pi_{\beta}\big{(}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\big{)}$.
Let

$t_{j}:=\pi_{\beta}\big{(}({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j})^{\infty}\big{)},\qquad
j=1,2,\dots,k-1.$

Note that $({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j})^{\infty}$ is
indeed the greedy $\beta$-expansion of $t_{j}$, since
$\mathbb{L}(({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j})^{\infty})\prec\mathbb{L}(({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k})^{\infty})\prec\alpha(\beta)$.
Noting also that ${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j}$ is
Lyndon, it follows that $T^{n}(t_{j})\geq t_{j}$ for every $n\geq 0$, so that
$t_{j}\in\mathscr{E}_{\beta}^{+}=\mathscr{E}_{\beta}$. Furthermore,
$({\mathbf{r}}_{1}\bullet\cdots\bullet{\mathbf{r}}_{j})^{\infty}\succ{\mathbf{r}}_{1}\bullet\cdots\bullet{\mathbf{r}}_{k}^{-}1^{\infty}\succ{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$.
Thus, $t_{j}\in(\tau(\beta),1)$.

It remains to show that there are no further points of $\mathscr{E}_{\beta}$
in $(\tau(\beta),1)$.

Recall that $k\geq 2$. We first claim that $K_{\beta}(t)=\emptyset$ for all
$t>t_{1}$, so there are no points of $\mathscr{E}_{\beta}$ to the right of
$t_{1}$. This follows since

$\mathcal{K}_{\beta}(t_{1})=\\{\mathbf{z}:{\mathbf{r}}_{1}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\prec\mathbb{L}({\mathbf{S}})^{\infty}\
\ \forall n\geq 0\\}=\Gamma({\mathbf{r}}_{1})$

by Lemma 8.2, so $\mathcal{K}_{\beta}(t_{1})$ consists only of
${\mathbf{r}}_{1}^{\infty}$ and its shifts by Lemma 6.2, and none of these lie
in $\mathcal{K}_{\beta}(t)$ for $t>t_{1}$.

Next, we show that there are no points of $\mathscr{E}_{\beta}$ in
$(\tau(\beta),t_{k-1})$; in other words, $K_{\beta}(t)$ is constant on
$(\tau(\beta),t_{k-1}]$. Note that

$\tau(\beta)=\pi_{\beta}\big{(}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\big{)}=\pi_{\beta}\big{(}{\mathbf{S}}0^{\infty}\big{)}=\pi_{\beta}\big{(}({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k})0^{\infty}\big{)}$

and
$t_{k-1}=\pi_{\beta}\big{(}({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1})^{\infty}\big{)}$.
Let $t\in(\tau(\beta),t_{k-1})$ and suppose
$\mathbf{z}\in\mathcal{K}_{\beta}(t)$. We claim that
$\mathbf{z}\in\mathcal{K}_{\beta}(t_{k-1})$. Observe that

(8.3) ${\mathbf{S}}0^{\infty}\prec
b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\prec\mathbb{L}({\mathbf{S}})^{\infty}\quad\forall
n\geq 0.$

Since ${\mathbf{S}}$ begins with
${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1}^{-}$, we have
$\sigma^{n}(\mathbf{z})\succcurlyeq({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1}^{-})0^{\infty}$
for all $n\geq 0$. Suppose $\mathbf{z}$ contains the word
${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1}^{-}$; without loss of
generality we may assume $\mathbf{z}$ begins with it. Then by Lemma 7.2,

$\mathbf{z}=\Phi_{{\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1}}(\hat{\mathbf{z}})$

for some sequence $\hat{\mathbf{z}}$ satisfying

${\mathbf{r}}_{k}0^{\infty}\preccurlyeq\sigma^{n}(\hat{\mathbf{z}})\prec\mathbb{L}({\mathbf{r}}_{k})^{\infty}\quad\forall
n\geq 0,$

in view of (8.3). However, by Lemma 6.2 such a sequence $\hat{\mathbf{z}}$
does not exist. Therefore,
$\sigma^{n}(\mathbf{z})\succcurlyeq({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{k-1})^{\infty}$
for all $n\geq 0$, which implies $\mathbf{z}\in\mathcal{K}_{\beta}(t_{k-1})$.

Note that we are now done with the case $k=2$, so fix $k\geq 3$ from now on.
Fix $1\leq j\leq k-2$, and let $t\in(t_{j+1},t_{j})$. Suppose by way of
contradiction that there exists a sequence
$\mathbf{z}\in\mathcal{K}_{\beta}(t)\backslash\mathcal{K}_{\beta}(t_{j})$.
Then

(8.4)
${\mathbf{S}}^{\infty}\preccurlyeq({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j+1})^{\infty}=b(t_{j+1},\beta)\prec
b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\prec\alpha(\beta)=\mathbb{L}({\mathbf{S}})^{\infty}\qquad\forall
n\geq 0,$

and there is an integer $n_{0}\geq 0$ such that
$\sigma^{n_{0}}(\mathbf{z})\prec
b(t_{j},\beta)=({\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j})^{\infty}$.
Since ${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j+1}$ begins with
${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j}^{-}$, there is therefore
a further integer $m_{0}\geq n_{0}$ such that $\sigma^{m_{0}}(\mathbf{z})$
begins with ${\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j}^{-}$. Set
${\mathbf{r}}:={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j}$ and
${\mathbf{s}}:={\mathbf{r}}_{j+1}\bullet\dots\bullet{\mathbf{r}}_{k}$.
Applying Lemma 7.2 with $\mathbf{x}={\mathbf{s}}^{\infty}$ and
$\mathbf{y}=\mathbb{L}({\mathbf{s}})^{\infty}$ we conclude that

$\mathbf{z}=\Phi_{{\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{j}}(\hat{\mathbf{z}})$

for some sequence $\hat{\mathbf{z}}$, which by (8.4) satisfies the
inequalities

(8.5)
${\mathbf{r}}_{j+1}^{\infty}\prec\sigma^{n}(\hat{\mathbf{z}})\prec\mathbb{L}({\mathbf{r}}_{j+1}\bullet\dots\bullet{\mathbf{r}}_{k})^{\infty}\qquad\forall
n\geq 0.$

But then

(8.6)
${\mathbf{r}}_{j+1}^{\infty}\prec\sigma^{n}(\hat{\mathbf{z}})\preccurlyeq\mathbb{L}({\mathbf{r}}_{j+1})^{\infty}\qquad\forall
n\geq 0$

by the obvious modification of Lemma 8.2. However, no sequence
$\hat{\mathbf{z}}$ satisfying (8.6) exists by Lemma 6.2. Hence,
$K_{\beta}(t)=K_{\beta}(t_{j})$.

This proves (i). The proof of (ii) is essentially the same, using that

$\tau(\beta_{*}^{\mathbf{S}})=\pi_{\beta}\big{(}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\big{)},\qquad\tau(\beta_{r}^{\mathbf{S}})=\pi_{\beta}\big{(}{\mathbf{S}}0^{\infty}\big{)}.$

The reason for the extra bifurcation point is that when
$\beta=\beta_{*}^{\mathbf{S}}$ or $\beta_{r}^{\mathbf{S}}$, the word
${\mathbf{S}}$ is actually $\beta$-Lyndon, whereas for
$\beta=\beta_{\ell}^{\mathbf{S}}$ it is not. ∎

##  9\. The case of infinitely renormalizable $\beta$

Recall the set $E_{\infty}$ from (2.11). In this section we prove the
following.

######  Theorem 9.1.

Let $\beta\in E_{\infty}$. Then

  1. (i)

For every $t_{R}\in\mathcal{T}_{R}$, $\widetilde{\mathcal{K}}_{\beta}(t_{R})$
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.

  2. (ii)

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

Part (i) gives Theorem 4.4 for $\beta\in E_{\infty}$.

###### Proof.

(i) Since $\beta\in E_{\infty}$, there is an infinite sequence
$\\{{\mathbf{s}}_{1},{\mathbf{s}}_{2},\dots\\}$ of Farey words, uniquely
determined by $\beta$, such that $\beta\in
J^{{\mathbf{s}}_{1}\bullet\dots\bullet{\mathbf{s}}_{n}}$ for each
$n\in\mathbb{N}$. Write
${\mathbf{S}}_{i}:={\mathbf{s}}_{1}\bullet\dots\bullet{\mathbf{s}}_{i}$ for
$i\in\mathbb{N}$, and note that for each $i$ there is a base $\hat{\beta}_{i}$
such that
$\alpha(\beta)=\Phi_{{\mathbf{S}}_{i}}\big{(}\alpha(\hat{\beta}_{i})\big{)}$.
Recall from [4, Proposition 6.3] that

(9.1)
$\tau(\beta)=\lim_{n\to\infty}\pi_{\beta}({\mathbf{s}}_{1}\bullet\cdots\bullet{\mathbf{s}}_{n}0^{\infty}).$

Hence, given a point $t_{R}\in\mathcal{T}_{R}$, we either have
$b(t_{R},\beta)\prec{\mathbf{s}}_{1}^{-}\mathbb{L}({\mathbf{s}}_{1})^{\infty}$,
in which case $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive by
Proposition 7.6, or else there is a unique $i\in\mathbb{N}$ such that

${\mathbf{S}}_{i}^{-}\mathbb{L}({\mathbf{S}}_{i})^{\infty}\prec
b(t_{R},\beta)\prec{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}.$

We conclude as in the proof of Proposition 7.8 that
$\widetilde{\mathcal{K}}_{\hat{\beta}_{i}}(\hat{t}_{R})$ is transitive. The
rest of the proof is the same as the proof of Proposition 7.8.

(ii) For each $n$, the $\beta$-Lyndon intervals are dense in $[0,\tau_{n}]$ by
applying the argument in Step 2 of the proof of Proposition 7.10 to
${\mathbf{S}}:={\mathbf{s}}_{1}\bullet\dots\bullet{\mathbf{s}}_{n}$. Since
$\tau_{n}\nearrow\tau(\beta)$, this implies the $\beta$-Lyndon intervals are
dense in $[0,\tau(\beta)]$. ∎

A simple modification of the proof of Theorem 8.1 yields the following:

######  Proposition 9.2.

Let $\beta\in E_{\infty}$ with corresponding sequence
${\mathbf{s}}_{1},{\mathbf{s}}_{2},\dots$ of Farey words. Then $\tau(\beta)$
is given by (9.1), and $\mathscr{E}_{\beta}$ has countably infinitely many
points in the interval $(\tau(\beta),1)$, namely the points whose greedy
$\beta$-expansions, in decreasing order, are

${\mathbf{s}}_{1}^{\infty},\qquad({\mathbf{s}}_{1}\bullet{\mathbf{s}}_{2})^{\infty},\qquad({\mathbf{s}}_{1}\bullet{\mathbf{s}}_{2}\bullet{\mathbf{s}}_{3})^{\infty},\qquad\dots.$

Unfortunately, the $\beta$-shift $\Sigma_{\beta}$ is never sofic for $\beta\in
E_{\infty}$, as it is easy to check that $\alpha(\beta)$ cannot be eventually
periodic. As a result, the subshifts $\widetilde{\mathcal{K}}_{\beta}(t_{R})$
are not all sofic, and therefore our argument from the previous section that
$\mathscr{E}_{\beta}\cap[0,\tau(\beta)]=\mathscr{B}_{\beta}\cap[0,\tau(\beta)]$
is not valid. We suspect, but have been unable to prove, that this equation is
nonetheless true, and hence $\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$
is countably infinite for $\beta\in E_{\infty}$.

##  10\. Interiors of basic intervals: Construction of non-transitivity
windows

We now begin to deal with the most complex case, which is when $\beta$ lies in
the interior of a basic interval. Before developing the general theory, we
give a concrete example which illustrates the main ideas.

######  Example 10.1.

Let $\beta$ be as in Example 5.10 such that $\alpha(\beta)$ begins with
$1110101100$, and set $\mathbf{w}=01010111$. For $t=t_{R}$ given by
$b(t_{R},\beta)=\mathbf{w}^{\infty}$, the subshift
$\widetilde{\mathcal{K}}_{\beta}(t)$ is not transitive: Note that, in fact,

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:(01010111)^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq(11101010)^{\infty}\
\forall\,n\geq 0\\},$

so the only sequence in $\widetilde{\mathcal{K}}_{\beta}(t)$ beginning with
$111$ is $(11101010)^{\infty}$, and the legal word $111$ cannot be connected
to, for instance, the legal word $011011$. On the other hand, disallowing the
word $111$ removes only countably many sequences from
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$, so the subshift

$\mathcal{K}_{\beta}^{\prime}(t):=\\{\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t):\mathbf{z}\
\mbox{does not contain $111$}\\}$

has full entropy in $\widetilde{\mathcal{K}}_{\beta}(t)$. Note that
$\mathcal{K}_{\beta}^{\prime}(t)$ consists precisely of those sequences of the
form

$(110)^{k_{1}}10(110)^{k_{2}}10(110)^{k_{3}}10\dots,\qquad k_{i}\geq 1\
\forall i,$

together with their orbits under $\sigma$, and is clearly transitive. Now let
$\mathbf{w}^{*}=01011$ and let $t^{*}$ be given by
$b(t^{*},\beta)=(\mathbf{w}^{*})^{\infty}$. It is easy to see that
$\widetilde{\mathcal{K}}_{\beta}(t^{*})=\mathcal{K}_{\beta}^{\prime}(t)$.
Hence,

$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t)\big{)}=h\big{(}\mathcal{K}_{\beta}^{\prime}(t)\big{)}=h\big{(}\widetilde{\mathcal{K}}_{\beta}(t^{*})\big{)}.$

Observe that $\mathbf{w}^{*}$ is the lexicographically smallest $\beta$-Lyndon
word greater than $\mathbf{w}$. Note also that
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ does not have a transitive subshift
of full entropy containing the sequence $\mathbf{w}^{\infty}$.

Developing the example a bit further, we set
$\mathbf{w}_{k}:=0101(011)^{k}0111$ for $k=0,1,2,\dots$, and let $t_{k}$ be
the point given by $b(t_{k},\beta)=\mathbf{w}_{k}^{\infty}$. As above it can
be shown that $\widetilde{\mathcal{K}}_{\beta}(t_{k})$ is not transitive,
though here there are more sequences in
$\widetilde{\mathcal{K}}_{\beta}(t_{k})$ beginning with $111$. With a bit more
effort, it can be shown that
$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{k})\big{)}=h\big{(}\widetilde{\mathcal{K}}_{\beta}(t^{*})\big{)}$.
Observe that $t_{k}\searrow t_{\infty}$, where $t_{\infty}$ is given by
$b(t_{\infty},\beta)=0101(011)^{\infty}$. By the continuity of $t\mapsto
h(\widetilde{\mathcal{K}}_{\beta}(t))$, it follows that
$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{\infty})\big{)}=h\big{(}\widetilde{\mathcal{K}}_{\beta}(t^{*})\big{)}$.

We call the interval $[t_{\infty},t^{*})$ a non-transitivity window. The
assertions about entropy in the last paragraph will follow from Proposition
11.6 below.

We saw in Section 5 that if $\beta\in E_{L}$, then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive for any $\beta$-Lyndon
interval $[t_{L},t_{R}]\subseteq[0,\tau(\beta)]$. On the other hand, the above
example shows that this need no longer be the case when $\beta$ lies in the
interior of a basic interval. We will now examine the question of transitivity
for the interior of basic intervals in greater detail. We will show that,
depending on $\beta$, there is a collection of intervals which we call non-
transitivity windows, such that $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is
transitive if and only if $t_{R}$ does not lie in any of these intervals. The
number of non-transitivity windows can be zero, positive and finite, or
infinite, depending on $\beta$.

We first prove a simple but useful lemma.

######  Lemma 10.2.

If a word $\mathbf{v}$ is Lyndon but not $\beta$-Lyndon, then no word
extending $\mathbf{v}$ is $\beta$-Lyndon.

###### Proof.

Suppose $\mathbf{v}=v_{1}\dots v_{k}$ is Lyndon but not $\beta$-Lyndon, so
there is an integer $j<k$ such that

(10.1) $v_{j+1}\dots v_{k}v_{1}\dots
v_{j}\succcurlyeq\alpha_{1}\dots\alpha_{k}.$

Let $\mathbf{w}=v_{1}\dots v_{k}v_{k+1}\dots v_{l}$ be an extension of
$\mathbf{v}$. Assume $\mathbf{w}$ is Lyndon, as otherwise there is nothing to
prove. Write $l=qk+r$, where $q\in\mathbb{N}$ and $0\leq r<k$. Since
$\mathbf{w}$ is Lyndon, $v_{ik+1}\dots v_{(i+1)k}\succcurlyeq v_{1}\dots
v_{k}$ for all $1\leq i<q$, with strict inequality for $i=q-1$ in case $r=0$;
and $v_{qk+1}\dots v_{l}=v_{qk+1}\dots v_{qk+r}\succ v_{1}\dots v_{r}$ if
$r>0$. Thus, with $j$ as in (10.1),

$v_{j+1}\dots v_{l}\succ v_{j+1}\dots v_{k}(v_{1}\dots v_{k})^{q-1}v_{1}\dots
v_{r}.$

If $q\geq 2$ or $r\geq j$, this gives

$(v_{j+1}\dots v_{l}v_{1}\dots v_{j})^{\infty}\succ v_{j+1}\dots
v_{l}0^{\infty}\succ(v_{j+1}\dots v_{k}v_{1}\dots
v_{j})^{\infty}\succcurlyeq\alpha(\beta).$

On the other hand, if $q=1$ and $r<j$, then $l-k=r<j$ and hence

$v_{j+1}\dots v_{l}\succ v_{j+1}\dots v_{k}v_{1}\dots
v_{l-k}\succcurlyeq\alpha_{1}\dots\alpha_{l-j},$

using (10.1). In both cases, it follows that $\mathbf{w}\not\in L^{*}(\beta)$.
∎

Now we fix a basic interval
$[\beta_{\ell},\beta_{*}]=[\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}}]$
generated by a word ${\mathbf{S}}\in\Lambda$, and fix
$\beta\in(\beta_{\ell},\beta_{*})$. Recall from [4, Theorem 2] that

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

Assume first that $\alpha(\beta)=\alpha_{1}\alpha_{2}\dots$ is not periodic.
We will inductively build a collection $\mathcal{I}$ of intervals in
$[0,\tau(\beta)]$ as follows.

To begin, set $j_{1}:=|{\mathbf{S}}|$. Then
$\sigma^{j_{1}}(\alpha(\beta))\prec\sigma^{j_{1}}(\alpha(\beta_{*}))={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$.
If $\sigma^{j_{1}+n}(\alpha(\beta))\succ\sigma^{j_{1}}(\alpha(\beta))$ for all
$n\geq 1$, then we set $\mathcal{I}:=\emptyset$. Otherwise, let

$n_{1}:=\min\\{n\geq
1:\sigma^{j_{1}+n}(\alpha(\beta))\preccurlyeq\sigma^{j_{1}}(\alpha(\beta))\\}.$

Then we set $\mathbf{v}_{1}:=\alpha_{j_{1}+1}\dots\alpha_{j_{1}+n_{1}}$, and
note that $\mathbf{v}_{1}$ is Lyndon (though not necessarily $\beta$-Lyndon).
To see this, take $1\leq i<n_{1}$ and observe by the definition of $n_{1}$
that

$\sigma^{j_{1}+n_{1}}(\alpha(\beta))\preccurlyeq\sigma^{j_{1}}(\alpha(\beta))\prec\sigma^{j_{1}+n_{1}-i}(\alpha(\beta)).$

Thus, by the minimality of $n_{1}$,

$\displaystyle\alpha_{j_{1}+1}\dots\alpha_{j_{1}+n_{1}-i}\sigma^{j_{1}+n_{1}-i}(\alpha(\beta))$
$\displaystyle=\sigma^{j_{1}}(\alpha(\beta))\prec\sigma^{j_{1}+i}(\alpha(\beta))$
$\displaystyle=\alpha_{j_{1}+i+1}\dots\alpha_{j_{1}+n_{1}}\sigma^{j_{1}+n_{1}}(\alpha(\beta))$
$\displaystyle\prec\alpha_{j_{1}+i+1}\dots\alpha_{j_{1}+n_{1}}\sigma^{j_{1}+n_{1}-i}(\alpha(\beta)),$

and this implies
$\alpha_{j_{1}+1}\dots\alpha_{j_{1}+n_{1}-i}\prec\alpha_{j_{1}+i+1}\dots\alpha_{n_{1}}$.

We define an interval $I_{1}=[\underline{t}_{1},\overline{t}_{1})$, where
$\underline{t}_{1}$ and $\overline{t}_{1}$ are given implicitly by

$b(\underline{t}_{1},\beta)=\mathbf{v}_{1}^{-}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{\infty}=\mathbf{v}_{1}^{-}\mathbb{L}({\mathbf{S}})^{\infty},\qquad
b(\overline{t}_{1},\beta)=(\mathbf{v}_{1}^{*})^{\infty},$

where for any Lyndon word $\mathbf{v}$, $\mathbf{v}^{*}$ denotes the
(lexicographically) smallest $\beta$-Lyndon word such that
$(\mathbf{v}^{*})^{\infty}\succcurlyeq\mathbf{v}^{\infty}$. (In particular,
$\mathbf{v}_{1}^{*}=\mathbf{v}_{1}$ if $\mathbf{v}_{1}$ is already
$\beta$-Lyndon.) Note by Lemma 10.2 that if $\mathbf{v}_{1}\not\in
L^{*}(\beta)$, then $\mathbf{v}_{1}^{*}$ is well defined and
$|\mathbf{v}_{1}^{*}|<|\mathbf{v}_{1}|$. We put the interval $I_{1}$ in the
collection $\mathcal{I}$.

Next, we set $j_{2}:=j_{1}+n_{1}$ and continue the procedure in the same way.
That is, if
$\sigma^{j_{2}+n}(\alpha(\beta))\succ\sigma^{j_{2}}(\alpha(\beta))$ for all
$n\geq 1$, then we set $\mathcal{I}=\\{I_{1}\\}$ and stop. Otherwise, let

$n_{2}:=\min\\{n\geq
1:\sigma^{j_{2}+n}(\alpha(\beta))\preccurlyeq\sigma^{j_{2}}(\alpha(\beta))\\},$

and set $\mathbf{v}_{2}:=\alpha_{j_{2}+1}\dots\alpha_{j_{2}+n_{2}}$. Observe
that $\mathbf{v}_{2}$ is Lyndon by the same argument that we used above for
$\mathbf{v}_{1}$. We define an interval
$I_{2}=[\underline{t}_{2},\overline{t}_{2})$, where $\underline{t}_{2}$ and
$\overline{t}_{2}$ are given implicitly by

$b(\underline{t}_{2},\beta)=\mathbf{v}_{2}^{-}(\alpha_{1}\dots\alpha_{j_{2}}^{-})^{\infty}=\mathbf{v}_{2}^{-}(\mathbb{L}({\mathbf{S}})\mathbf{v}_{1}^{-})^{\infty},\qquad
b(\overline{t}_{2},\beta)=(\mathbf{v}_{2}^{*})^{\infty}.$

We add $I_{2}$ to the collection $\mathcal{I}$.

This way we continue: Suppose integers $j_{1},\dots,j_{k}$ and
$n_{1},\dots,n_{k}$, words $\mathbf{v}_{1},\dots,\mathbf{v}_{k}$ and intervals
$I_{1}=[\underline{t}_{1},\overline{t}_{1}),\dots,I_{k}=[\underline{t}_{k},\overline{t}_{k})$
have been constructed, and set $j_{k+1}:=j_{k}+n_{k}$. If
$\sigma^{j_{k+1}+n}(\alpha(\beta))\succ\sigma^{j_{k+1}}(\alpha(\beta))$ for
all $n\geq 1$, then we set $\mathcal{I}=\\{I_{1},\dots,I_{k}\\}$ and stop.
Otherwise, let

$n_{k+1}:=\min\\{n\geq
1:\sigma^{j_{k+1}+n}(\alpha(\beta))\preccurlyeq\sigma^{j_{k+1}}(\alpha(\beta))\\},$

and set $\mathbf{v}_{k+1}:=\alpha_{j_{k+1}+1}\dots\alpha_{j_{k+1}+n_{k+1}}$.
We define an interval $I_{k+1}=[\underline{t}_{k+1},\overline{t}_{k+1})$,
where $\underline{t}_{k+1}$ and $\overline{t}_{k+1}$ are given implicitly by

$\displaystyle b(\underline{t}_{k+1},\beta)$
$\displaystyle=\mathbf{v}_{k+1}^{-}(\alpha_{1}\dots\alpha_{j_{k+1}}^{-})^{\infty}=\mathbf{v}_{k+1}^{-}\big{(}\mathbb{L}({\mathbf{S}})\mathbf{v}_{1}\dots\mathbf{v}_{k-1}\mathbf{v}_{k}^{-}\big{)}^{\infty},$
$\displaystyle b(\overline{t}_{k+1},\beta)$
$\displaystyle=(\mathbf{v}_{k+1}^{*})^{\infty},$

However, in the special case when $\mathbf{v}_{k+1}\in L^{*}(\beta)$ and
$\sigma^{j_{k+1}}(\alpha(\beta))=\mathbf{v}_{k+1}^{\infty}$, we take the
interval $I_{k+1}$ to be closed, i.e.
$I_{k+1}=[\underline{t}_{k+1},\overline{t}_{k+1}]$. We add $I_{k+1}$ to the
collection $\mathcal{I}$.

This procedure either eventually stops, with a finite collection
$\mathcal{I}=\\{I_{1},\dots,I_{k_{0}}\\}$, or it continues ad infinitum,
producing an infinite collection $\mathcal{I}=\\{I_{1},I_{2},\dots\\}$ of
intervals. Note that if $I_{k}$ is closed for some $k$, then
$\mathbf{v}_{l}=\mathbf{v}_{k}$ for all $l>k$ and hence $I_{l}\subseteq I_{k}$
for all $l>k$.

In case $\alpha(\beta)$ is periodic, i.e.
$\alpha(\beta)=(\alpha_{1}\dots\alpha_{m})^{\infty}$ for some $m\in\mathbb{N}$
such that $\alpha_{m}=0$, we replace $(\alpha_{i})$ in the above procedure
with the greedy expansion of $1$, that is, the sequence
$(\alpha_{i}^{\prime}):=\alpha_{1}\dots\alpha_{m-1}10^{\infty}$. The inductive
step needs to be slightly modified: If
$\sigma^{j_{k+1}}((\alpha_{i}^{\prime}))=0^{\infty}$, we leave $n_{k+1}$
undefined and stop the construction with the finite set
$\mathcal{I}=\\{I_{1},\dots,I_{k}\\}$.

The following special case of Theorem 4.4 is our main result in this section.

######  Theorem 10.3.

Let $\beta\in(\beta_{\ell},\beta_{*})$ for some basic interval
$[\beta_{\ell},\beta_{*}]=[\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}]$,
where ${\mathbf{S}}\in\Lambda_{1}=F^{*}$, and construct the collection
$\mathcal{I}$ of intervals as above. Let
$[t_{L},t_{R}]\subseteq[0,\tau(\beta)]$ be a $\beta$-Lyndon interval. Then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive if and only if
$t_{R}\not\in\bigcup_{I\in\mathcal{I}}I$. Furthermore, the entropy of
$\widetilde{\mathcal{K}}_{\beta}(t)$ is constant on each interval
$I\in\mathcal{I}$.

The situation for ${\mathbf{S}}\in\Lambda_{k}$ with $k\geq 2$ is a bit more
involved, and we deal with it later, in Section 12. However, it is convenient
to develop the necessary machinery in greater generality here.

Before proving the theorem, which shows the significance of the words
$\mathbf{v}_{k}$ and intervals $I_{k}$, we first give some examples.

######  Example 10.4.

Let ${\mathbf{S}}=011$, so ${\mathbf{S}}\in F^{*}$ and the basic interval
$[\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}}]$ is given by

$\alpha(\beta_{\ell}^{{\mathbf{S}}})=\mathbb{L}({\mathbf{S}})^{\infty}=(110)^{\infty},\qquad\alpha(\beta_{*}^{{\mathbf{S}}})=\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=111\,010\,(110)^{\infty}.$

  1. (a)

Suppose $\alpha(\beta)=111\,01\,00110111\,(001)^{\infty}$. Then
$\mathbf{v}_{1}=01$, $\mathbf{v}_{2}=00110111$, and $\mathbf{v}_{k}=001$ for
all $k\geq 3$. Since each $\mathbf{v}_{k}\in L^{*}(\beta)$, we have
$\mathbf{v}_{k}^{*}=\mathbf{v}_{k}$ for each $k$. Now our construction above
gives rise to the intervals
$I_{1}=[\underline{t}_{1},\overline{t}_{1}),I_{2}=[\underline{t}_{2},\overline{t}_{2}),\dots$
given by

$\displaystyle b(\underline{t}_{1},\beta)$
$\displaystyle=\mathbf{v}_{1}^{-}\mathbb{L}({\mathbf{S}})^{\infty}=00\,(110)^{\infty},$
$\displaystyle\quad b(\overline{t}_{1},\beta)$
$\displaystyle=\mathbf{v}_{1}^{\infty}=(01)^{\infty},$ $\displaystyle
b(\underline{t}_{1},\beta)$
$\displaystyle=\mathbf{v}_{2}^{-}\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}^{-}\big{)}^{\infty}=00110110\,(11100)^{\infty},$
$\displaystyle\quad b(\overline{t}_{2},\beta)$
$\displaystyle=\mathbf{v}_{2}^{\infty}=(00110111)^{\infty},$ $\displaystyle
b(\underline{t}_{3},\beta)$
$\displaystyle=\mathbf{v}_{3}^{-}\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\mathbf{v}_{2}^{-}\big{)}^{\infty}=000\,(111\,01\,00110110)^{\infty},$
$\displaystyle\quad b(\overline{t}_{3},\beta)$
$\displaystyle=\mathbf{v}_{3}^{\infty}=(001)^{\infty},$

and for $k\geq 4$,

$\displaystyle b(\underline{t}_{k},\beta)$
$\displaystyle=\mathbf{v}_{k}^{-}\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k-1}^{-}\big{)}^{\infty}=000\,\big{(}111\,01\,00110111\,(001)^{k-4}\,000\big{)}^{\infty},$
$\displaystyle b(\overline{t}_{k},\beta)$
$\displaystyle=\mathbf{v}_{k}^{\infty}=(001)^{\infty}.$

Observe that $I_{2}\subseteq I_{1}$, $I_{3}$ lies to the left of $I_{2}$, and
$I_{k+1}\subseteq I_{k}$ for all $k\geq 3$.

  2. (b)

Suppose $\alpha(\beta)=111\,01\,001(01)^{\infty}$. Then $\mathbf{v}_{1}=01$,
and the procedure stops after one step, giving rise to just one interval
$I_{1}=[\underline{t}_{1},\overline{t}_{1})$, which is the same as in (a).

  3. (c)

More extremely, let $\alpha(\beta)=111\,001(01)^{\infty}$. Then
$\sigma^{j_{1}+n}(\alpha(\beta))\succ\sigma^{j_{1}}(\alpha(\beta))$ for all
$n\geq 1$ (where $j_{1}=|{\mathbf{S}}|=3$), and hence we obtain the empty
collection $\mathcal{I}=\emptyset$.

  4. (d)

Let $\alpha(\beta)=111\,01\,001\,0001\dots$. Then $\mathbf{v}_{k}=0^{k}1$ for
$k\in\mathbb{N}$, and we get an infinite collection
$\mathcal{I}=\\{I_{1},I_{2},\dots\\}$ of pairwise disjoint intervals, with
left endpoints given by

$\displaystyle b(\underline{t}_{1},\beta)=00\,(110)^{\infty},\qquad
b(\underline{t}_{2},\beta)=000\,(11100)^{\infty},$ $\displaystyle
b(\underline{t}_{k},\beta)=0^{k+1}\,\big{(}111\,01\,001\,\dots\,0^{k-2}1\,0^{k}\big{)}^{\infty}\qquad(k\geq
3),$

and right endpoints given by

$b(\overline{t}_{k},\beta)=(0^{k}1)^{\infty},\qquad k=1,2,\dots.$

Note that in this example, the intervals $I_{k}$ accumulate at $0$.

  5. (e)

Suppose $\alpha(\beta)=111\,00111\,(001)^{\infty}$. Then
$\mathbf{v}_{1}=00111$ and $\mathbf{v}_{k}=001$ for all $k\geq 2$. Note that
here $\mathbf{v}_{1}\not\in L^{*}(\beta)$ since
$\sigma^{2}(\mathbf{v}_{1}^{\infty})=(11100)^{\infty}\succ\alpha(\beta)$. It
is easy to check that $\mathbf{v}_{1}^{*}=01$. Thus, we get a collection of
intervals
$I_{1}=[\underline{t}_{1},\overline{t}_{1}),I_{2}=[\underline{t}_{2},\overline{t}_{2}),\dots$
given by

$\displaystyle b(\underline{t}_{1},\beta)$
$\displaystyle=00110\,(110)^{\infty},$ $\displaystyle\qquad
b(\overline{t}_{1},\beta)$ $\displaystyle=(01)^{\infty},$ $\displaystyle
b(\underline{t}_{2},\beta)$ $\displaystyle=000\,(111\,00110)^{\infty},$
$\displaystyle\qquad b(\overline{t}_{2},\beta)$
$\displaystyle=(001)^{\infty},$ $\displaystyle b(\underline{t}_{k},\beta)$
$\displaystyle=000\,\big{(}111\,00111\,(001)^{k-3}\,000\big{)}^{\infty},$
$\displaystyle\qquad b(\overline{t}_{k},\beta)$
$\displaystyle=(001)^{\infty}\qquad(k\geq 3).$

  6. (f)

Let $\alpha(\beta)=(111\,0101100)^{\infty}$. Since $\alpha(\beta)$ is
periodic, we apply the construction to the greedy expansion
$\alpha_{1}^{\prime}\alpha_{2}^{\prime}\dots=111\,01011\,01\,0^{\infty}$. We
find $\mathbf{v}_{1}=01011$ and $\mathbf{v}_{2}=01$. The intervals
$I_{1}=[\underline{t}_{1},\overline{t}_{1})$ and
$I_{2}=[\underline{t}_{2},\overline{t}_{2})$ are given by

$\displaystyle b(\underline{t}_{1},\beta)$
$\displaystyle=01010\,(110)^{\infty},$ $\displaystyle\qquad
b(\overline{t}_{1},\beta)$ $\displaystyle=(01011)^{\infty},$ $\displaystyle
b(\underline{t}_{2},\beta)$ $\displaystyle=00\,(111\,01010)^{\infty},$
$\displaystyle\qquad b(\overline{t}_{2},\beta)$ $\displaystyle=(01)^{\infty}.$

The proof of Theorem 10.3 is rather involved. We first prove several lemmas
about the words $\mathbf{v}_{k}$. These lemmas lead to Proposition 10.13,
which describes the relative placements of the intervals $I_{k}$. The first
lemma deals specifically with the case of periodic $\alpha(\beta)$; see [22,
Lemma 3.8].

######  Lemma 10.5.

Let $\beta\in(1,2)$, and suppose
$\alpha(\beta)=(\alpha_{1}\dots\alpha_{m})^{\infty}$ where $m$ is the minimal
period of $\alpha(\beta)$. Then

$\alpha_{j+1}\dots\alpha_{m}\prec\alpha_{1}\dots\alpha_{m-j}\qquad\forall\,1\leq
j<m.$

Below we write $\mathbf{v}_{k}=v_{k,1}\dots v_{k,n_{k}}$. Recall also that
$n_{k}$ is the length of $\mathbf{v}_{k}$.

######  Lemma 10.6.

For any $k$ and any two integers $0\leq i_{1}<i_{2}\leq n_{k}$,

(10.2) $v_{k,i_{1}+1}\dots
v_{k,i_{2}}\preccurlyeq\alpha_{1}\dots\alpha_{i_{2}-i_{1}}.$

###### Proof.

When $\alpha(\beta)$ is aperiodic the inequality follows immediately from
Lemma 2.1, since $\mathbf{v}_{k}$ is a subword of $\alpha(\beta)$. But in the
periodic case some extra care is needed. In this case there is a positive
integer $k_{0}$ such that $\mathcal{I}=\\{I_{1},\dots,I_{k_{0}}\\}$, and
$\alpha(\beta)=(\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-})^{\infty}$.
The only case that needs extra attention is when $k=k_{0}$ and
$i_{2}=n_{k_{0}}$. Set $i:=i_{1}$. It seems theoretically possible that
$v_{k_{0},i+1}\dots
v_{k_{0},n_{k_{0}}}=\alpha_{1}\dots\alpha_{n_{k_{0}}-i}^{+}$, violating
(10.2). However, if that were the case we would have

$\alpha_{1}\dots\alpha_{m}=\alpha_{1}\dots\alpha_{j_{k_{0}}}\mathbf{v}_{k_{0}}^{-}=\alpha_{1}\dots\alpha_{j_{k_{0}}}v_{k_{0},1}\dots
v_{k_{0},i}\alpha_{1}\dots\alpha_{n_{k_{0}}-i},$

where $m$ is the minimal period of $\alpha(\beta)$, so applying Lemma 10.5
with $j=n_{k_{0}}-i$ gives

$\sigma^{j_{k_{0}}+i}(\alpha(\beta))=\alpha_{1}\dots\alpha_{n_{k_{0}}-i}(\alpha_{1}\dots\alpha_{m})^{\infty}\succ\alpha(\beta).$

This contradiction shows that (10.2) holds for this case as well. ∎

######  Lemma 10.7.

Let $k\geq 2$. If $\mathbf{v}_{k}$ is a proper prefix of $\mathbf{v}_{k-1}$,
then $\mathbf{v}_{k}\in L^{*}(\beta)$.

###### Proof.

By the hypothesis, $n_{k}<n_{k-1}$ and we can write
$\mathbf{v}_{k}=v_{k-1,1}\dots v_{k-1,n_{k}}$. Fix $0\leq i<n_{k}$. By (10.2),
$v_{k-1,i+1}\dots v_{k-1,n_{k}}\preccurlyeq\alpha_{1}\dots\alpha_{n_{k}-i}$.
Write $n_{k-1}=qn_{k}+r$, where $q\in\mathbb{N}$ and $1\leq r\leq n_{k}$.
Since $\mathbf{v}_{k-1}$ is Lyndon, we have

$\mathbf{v}_{k}=v_{k-1,1}\dots v_{k-1,n_{k}}\preccurlyeq v_{k-1,pn_{k}+1}\dots
v_{k-1,(p+1)n_{k}},\qquad p=0,1,\dots,q-1,$

and

$v_{k-1,1}\dots v_{k-1,r}\prec v_{k-1,qn_{k}+1}\dots v_{k-1,n_{k-1}}.$

Thus, by Lemma 10.6 it follows that

$v_{k-1,i+1}\dots v_{k-1,n_{k}}\mathbf{v}_{k}^{q-1}v_{k-1,1}\dots
v_{k-1,r}\prec v_{k-1,i+1}\dots v_{k-1,n_{k-1}}\preccurlyeq\alpha_{1}\dots
a_{n_{k-1}-i},$

and therefore, $\sigma^{i}(\mathbf{v}_{k}^{\infty})\prec\alpha(\beta)$. Since
$i$ was arbitrary, this shows $\mathbf{v}_{k}\in L^{*}(\beta)$. ∎

######  Lemma 10.8.

Let $k\geq 2$, and suppose $\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. Then

  1. (i)

$\mathbf{v}_{k-1}$ is not a prefix of $\mathbf{v}_{k}$; and

  2. (ii)

$\mathbf{v}_{k}^{\infty}\prec\mathbf{v}_{k-1}^{\infty}$.

###### Proof.

Since
$\sigma^{j_{k}}(\alpha(\beta))=\sigma^{j_{k-1}+n_{k-1}}(\alpha(\beta))\preccurlyeq\sigma^{j_{k-1}}(\alpha(\beta))$
and $\mathbf{v}_{k-1}$ is a prefix of $\sigma^{j_{k-1}}(\alpha(\beta))$, it
follows that

$\mathbf{v}_{k}\sigma^{j_{k}+n_{k}}(\alpha(\beta))=\sigma^{j_{k}}(\alpha(\beta))\preccurlyeq\mathbf{v}_{k-1}^{\infty},$

as is easy to check. Furthermore, since $\mathbf{v}_{k-1}$ and
$\mathbf{v}_{k}$ are both Lyndon, $\mathbf{v}_{k}$ cannot extend
$\mathbf{v}_{k-1}$. This yields (i), and also yields (ii) in case
$\mathbf{v}_{k}$ is not a prefix of $\mathbf{v}_{k-1}$. If $\mathbf{v}_{k}$ is
a prefix of $\mathbf{v}_{k-1}$, then $\mathbf{v}_{k}$ is a proper prefix of
$\mathbf{v}_{k-1}$ by our assumption $\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. So
we can write $n_{k-1}=qn_{k}+r$, where $q\in\mathbb{N}$ and $1\leq r\leq
n_{k}$. By a similar argument as in the proof of Lemma 10.7, the Lyndon
property of $\mathbf{v}_{k-1}$ implies
$\mathbf{v}_{k-1}\succ\mathbf{v}_{k}^{q}v_{k,1}\dots v_{k,r},$ proving (ii). ∎

######  Lemma 10.9.

For each $k\in\mathbb{N}$, the word $\alpha_{1}\dots\alpha_{j_{k}}$ can occur
in $\mathbf{v}_{k}$ at most once, and only at the end. In other words, the
truncated word $v_{k,1}\dots v_{k,n_{k}-1}$ does not contain the word
$\alpha_{1}\dots\alpha_{j_{k}}$.

###### Proof.

Suppose that $\mathbf{v}_{k}=v_{k,1}\dots
v_{k,p}\alpha_{1}\dots\alpha_{j_{k}}u_{1}\dots u_{q}$ (say), where $p\geq 0$
and $q\geq 1$. Then, on the one hand, $u_{1}\dots u_{q}\succ v_{k,1}\dots
v_{k,q}$ since $\mathbf{v}_{k}$ is Lyndon. On the other hand, applying (10.2)
with $i_{1}=p$ and $i_{2}=n_{k}$ we obtain $u_{1}\dots
u_{q}\preccurlyeq\alpha_{j_{k}+1}\dots\alpha_{j_{k}+q}=v_{k,1}\dots v_{k,q}$.
This contradiction completes the proof. ∎

######  Lemma 10.10.

  1. (i)

For each $k\in\mathbb{N}$, we have

(10.3)
$\mathbf{v}_{k}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{k}}\qquad\forall\,0\leq
m<j_{k}.$

  2. (ii)

Suppose either $k=1$, or $k\geq 2$ and $\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$.
If $\mathbf{v}_{k}\not\in L^{*}(\beta)$, then

(10.4)
$\mathbf{v}_{k}\prec\alpha_{m+1}\dots\alpha_{m+n_{k}}\qquad\forall\,0\leq
m<j_{k}.$

###### Proof.

We prove the two statements by induction on $k$. Take first $k=1$. We first
show that

(10.5)
$\mathbf{v}_{1}\preccurlyeq({\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty})_{1:n_{1}}.$

(Here, for a word $\mathbf{u}=u_{1}\dots u_{n}$ or sequence
$\mathbf{u}=u_{1}u_{2}\dots$, we use the notation
$\mathbf{u}_{i:j}:=u_{i}\dots u_{j}$.) Recall that $j_{1}=|{\mathbf{S}}|$, and
$\sigma^{j_{1}}(\alpha(\beta))\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$.
Furthermore,
$\alpha(\beta)\succcurlyeq(\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}^{-})^{\infty}$,
with equality possible in the periodic case. Thus,

${\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\succ\sigma^{j_{1}}(\alpha(\beta))\succcurlyeq(\mathbf{v}_{1}^{-}\mathbb{L}({\mathbf{S}})^{+})^{\infty}.$

This shows that
$\mathbf{v}_{1}^{-}\preccurlyeq({\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty})_{1:n_{1}}$.
If this holds with equality, then we obtain

$\sigma^{n_{1}}({\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty})\succ(\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}^{-})^{\infty},$

which is impossible since ${\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$
does not contain the word $\mathbb{L}({\mathbf{S}})^{+}$. Hence, we have
(10.5).

It follows that, for any $m<j_{1}$,

(10.6)
$\displaystyle\begin{split}\mathbf{v}_{1}&\preccurlyeq({\mathbf{S}}0^{\infty})_{1:n_{1}}\preccurlyeq(\alpha_{m+1}\dots\alpha_{j_{1}}^{-}\alpha_{1}\dots\alpha_{m}0^{\infty})_{1:n_{1}}\\\
&\preccurlyeq(\alpha_{m+1}\dots\alpha_{j_{1}}0^{\infty})_{1:n_{1}}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{1}},\end{split}$

where the second inequality holds because
$\mathbb{L}({\mathbf{S}})=\alpha_{1}\dots\alpha_{j_{1}}^{-}$, and so
$\alpha_{m+1}\dots\alpha_{j_{1}}^{-}\alpha_{1}\dots\alpha_{m}$ is a cyclic
permutation of ${\mathbf{S}}$. Observe that the third inequality in (10.6) is
strict when $n_{1}\geq j_{1}-m$.

Now suppose that $\mathbf{v}_{1}\not\in L^{*}(\beta)$, and $n_{1}<j_{1}-m$.
Observe that ${\mathbf{S}}\in L^{*}(\beta)$ since
$\sigma^{n}({\mathbf{S}}^{\infty})\preccurlyeq\mathbb{L}({\mathbf{S}})^{\infty}\prec\mathbb{L}({\mathbf{S}})^{+}0^{\infty}\prec\alpha(\beta)$
for each $n\geq 0$. Thus, by Lemma 10.2, $\mathbf{v}_{1}$ cannot be a prefix
of ${\mathbf{S}}$. But $|\mathbf{v}_{1}|=n_{1}<j_{1}=|{\mathbf{S}}|$, and
hence $\mathbf{v}_{1}\prec s_{1}\dots
s_{n_{1}}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{1}}$ as in (10.6). This
completes the basis for the induction.

Next, let $k\geq 2$ be given, and suppose that
$\mathbf{v}_{k-1}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{k-1}}$ for all
$m<j_{k-1}$. Take $m<j_{k}$, and consider two cases:

Case 1. $m\leq j_{k-1}$. If $n_{k}\leq n_{k-1}$, then Lemma 10.8 (ii) and the
induction hypothesis imply that $\mathbf{v}_{k}\preccurlyeq v_{k-1,1}\dots
v_{k-1,n_{k}}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{k}}$, giving (10.3).
Furthermore, the first inequality is strict in case
$\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$ and $\mathbf{v}_{k}\not\in L^{*}(\beta)$,
in view of Lemma 10.7, so in this case we have (10.4). If, on the other hand,
$n_{k}>n_{k-1}$, then $v_{k,1}\dots
v_{k,n_{k-1}}\prec\mathbf{v}_{k-1}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+n_{k-1}}$,
again by the induction hypothesis, where the first inequality follows from
Lemma 10.8 (i).

Case 2. $m>j_{k-1}$. Then $j_{k}-m<j_{k}-j_{k-1}=n_{k-1}$, and we use the
Lyndon property of $\mathbf{v}_{k-1}$ to argue as follows: If $j_{k}-m\leq
n_{k}$, then

$v_{k,1}\dots v_{k,j_{k}-m}\preccurlyeq v_{k-1,1}\dots v_{k-1,j_{k}-m}\prec
v_{k-1,m-j_{k-1}+1}\dots v_{k-1,n_{k-1}}=\alpha_{m+1}\dots\alpha_{j_{k}},$

which implies (10.4); and if $j_{k}-m>n_{k}$, then

$\mathbf{v}_{k}\prec v_{k-1,m-j_{k-1}+1}\dots
v_{k-1,m+n_{k}-j_{k-1}}=\alpha_{m+1}\dots\alpha_{m+n_{k}},$

using that $\mathbf{v}_{k}$ does not extend $\mathbf{v}_{k-1}$ by Lemma 10.8
(i). ∎

######  Lemma 10.11.

Let $k\in\mathbb{N}$, and suppose either $k=1$, or $k\geq 2$ and
$\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. Then $\mathbf{v}_{k}\in L^{*}(\beta)$ if
and only if $\mathbf{v}_{k}$ does not end in $\alpha_{1}\dots\alpha_{j_{k}}$.

###### Proof.

If $\mathbf{v}_{k}=\mathbf{u}\alpha_{1}\dots\alpha_{j_{k}}$ for some word
$\mathbf{u}$, then

$\sigma^{|\mathbf{u}|}(\mathbf{v}_{k}^{\infty})=(\alpha_{1}\dots\alpha_{j_{k}}\mathbf{u})^{\infty}=(\alpha_{1}\dots\alpha_{j_{k}+|\mathbf{u}|})^{\infty}\succcurlyeq\alpha(\beta),$

and hence $\mathbf{v}_{k}\not\in L^{*}(\beta)$.

Vice versa, suppose $\mathbf{v}_{k}\not\in L^{*}(\beta)$. Then there is an
index $i\leq n_{k}$ such that

(10.7) $(v_{k,i+1}\dots v_{k,n_{k}}v_{k,1}\dots
v_{k,i})^{\infty}=v_{k,i+1}\dots
v_{k,n_{k}}\mathbf{v}_{k}^{\infty}\succcurlyeq\alpha(\beta).$

In view of (10.2), this implies

(10.8) $v_{k,i+1}\dots v_{k,n_{k}}=\alpha_{1}\dots\alpha_{n_{k}-i}.$

We claim that $n_{k}-i=j_{k}$, and so $\mathbf{v}_{k}$ ends in
$\alpha_{1}\dots\alpha_{j_{k}}$. First, if $n_{k}-i>j_{k}$, then (10.8) gives

$v_{k,i+j_{k}+1}\dots
v_{k,n_{k}}=\alpha_{j_{k}+1}\dots\alpha_{n_{k}-i}=v_{k,1}\dots
v_{k,n_{k}-i-j_{k}},$

contradicting that $\mathbf{v}_{k}$ is Lyndon. On the other hand, if
$n_{k}-i<j_{k}$, then
$\mathbf{v}_{k}\prec\alpha_{n_{k}-i+1}\dots\alpha_{2n_{k}-i}$ by Lemma 10.10
(ii), and together with (10.8), this contradicts (10.7).

Thus, $n_{k}-i=j_{k}$, and the proof is complete. ∎

We are now ready to characterize $\mathbf{v}_{k}^{*}$.

######  Lemma 10.12.

Let $k\geq 1$, and suppose $\mathbf{v}_{k}\not\in L^{*}(\beta)$. If $k\geq 2$
and $\mathbf{v}_{k}=\mathbf{v}_{k-1}$, then
$\mathbf{v}_{k}^{*}=\mathbf{v}_{k-1}^{*}$. Otherwise, there exist a
nonnegative integer $r$ and a word $\mathbf{u}$ not ending in
$\alpha_{1}\dots\alpha_{j_{k}}^{-}$ such that

(10.9)
$\mathbf{v}_{k}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}},$

and we have

(10.10) $\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}$

###### Proof.

Assume that either $k=1$, or $k\geq 2$ and
$\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. Then by the previous lemma,
$\mathbf{v}_{k}$ ends in $\alpha_{1}\dots\alpha_{j_{k}}$, and this implies the
existence of the word $\mathbf{u}$ and integer $r\geq 0$ satisfying (10.9).
Before determining $\mathbf{v}_{k}^{*}$, we first show that $\mathbf{u}$
cannot end in a prefix of $\alpha(\beta)$. For, suppose $\mathbf{u}$ ends in
$\alpha_{1}\dots\alpha_{m}$ for some $m\geq 1$. We claim that this contradicts
Lemma 10.9. Note by (10.9) and (10.2) that

$(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+(r+1)j_{k}}\preccurlyeq\alpha_{1}\ldots\alpha_{(r+1)j_{k}}.$

This implies that
$\alpha_{m+1}\dots\alpha_{m+j_{k}}=\alpha_{1}\dots\alpha_{j_{k}}$ or
$\alpha_{1}\dots\alpha_{j_{k}}^{-}$. By (10.9) it follows that
$\mathbf{v}_{k}$ contains the word
$\alpha_{1}\dots\alpha_{m}\alpha_{m+1}\dots\alpha_{m+j_{k}-1}$, followed by
one more digit. Since $m\geq 1$, this means that $\mathbf{v}_{k}$ contains the
word $\alpha_{1}\dots\alpha_{j_{k}}$ prematurely, contradicting Lemma 10.9.

We next prove the expression (10.10). First, let $k=1$. We need to show that
$\mathbf{v}_{1}^{*}=\mathbf{u}^{+}$. Note that
$(\mathbf{v}_{1}^{*})^{\infty}\succ\mathbf{v}_{1}^{\infty}$ and
$|\mathbf{v}_{1}^{*}|<|\mathbf{v}_{1}|$ by Lemma 10.2. First we claim that
$\mathbf{u}$ cannot be a prefix of $\mathbf{v}_{1}^{*}$. If $r=0$, then
$\mathbf{v}_{1}=\mathbf{u}\alpha_{1}\dots\alpha_{j_{1}}$ by (10.9), and since
$|\mathbf{v}_{1}^{*}|<|\mathbf{v}_{1}|$, the constraint (10.2) implies that
$|\mathbf{v}_{1}^{*}|<|\mathbf{u}|$, with strict inequality because
$\mathbf{u}$ ends in $0$ and $\mathbf{v}_{1}^{*}$ ends in $1$.

If $r\geq 1$, we need to rule out the possibility that $\mathbf{v}_{1}^{*}$
might equal
$\mathbf{w}:=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{p}\alpha_{1}\dots\alpha_{j_{1}}$
for some $p<r$. But notice that

(10.11)
$\sigma^{|\mathbf{u}|+pj_{1}}(\mathbf{w}^{\infty})=\alpha_{1}\dots\alpha_{j_{1}}\mathbf{w}^{\infty}\succ\alpha_{1}\dots\alpha_{j_{1}}\mathbf{v}_{1}^{\infty}=(\alpha_{1}\dots\alpha_{n_{1}})^{\infty}\succcurlyeq\alpha(\beta),$

where the second equality holds since $\mathbf{v}_{1}$ ends in
$\alpha_{1}\dots\alpha_{j_{1}}$, so $n_{1}>j_{1}$ and
$\mathbf{v}_{1}=\alpha_{j_{1}+1}\dots\alpha_{j_{1}+n_{1}}=\alpha_{j_{1}+1}\dots\alpha_{n_{1}}\alpha_{1}\dots\alpha_{j_{1}}$.
As a consequence of (10.11), $\mathbf{w}\not\in L^{*}(\beta)$, and so
$\mathbf{u}$ cannot be a prefix of $\mathbf{v}_{1}^{*}$. This proves the
claim.

It now suffices to show that $\mathbf{u}^{+}\in L^{*}(\beta)$. Note
$\mathbf{u}^{+}$ is Lyndon because $\mathbf{u}$ is a prefix of
$\mathbf{v}_{1}$ and $\mathbf{v}_{1}$ is Lyndon. Note that
$j_{1}=|{\mathbf{S}}|$. Write ${\mathbf{S}}=s_{1}\dots s_{j_{1}}$ and
$\mathbf{u}=u_{1}\dots u_{l}$. If $l\leq j_{1}$ and $\mathbf{u}=s_{1}\dots
s_{l}^{-}$, then $\mathbf{u}^{+}=s_{1}\dots s_{l}\in L^{*}(\beta)$ by Lemma
10.2, because ${\mathbf{S}}\in L^{*}(\beta)$. So assume $\mathbf{u}$ is not of
this form. We showed above that $\mathbf{u}$ cannot end in a prefix of
$\alpha(\beta)$, so if $\mathbf{u}^{+}\not\in L^{*}(\beta)$, then $\mathbf{u}$
must end in $\alpha_{1}\dots\alpha_{m}^{-}$ for some $m\geq 1$, i.e.
$\mathbf{u}=u_{1}\dots u_{l-m}\alpha_{1}\dots\alpha_{m}^{-}$. Observe that
$m\leq j_{1}$ by Lemma 10.9, and $m\neq j_{1}$ because we assumed that
$\mathbf{u}$ does not end in $\alpha_{1}\dots\alpha_{j_{1}}^{-}$. Thus,
$m<j_{1}$.

We claim next that $\mathbf{u}$ is not a prefix of ${\mathbf{S}}^{-}$. Suppose
it is, so $\mathbf{u}=s_{1}\dots s_{l}$ where $l<j_{1}$. Then

$\mathbf{v}_{1}^{-}=s_{1}\dots
s_{l}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r+1}=s_{1}\dots
s_{l}\mathbb{L}({\mathbf{S}})^{r+1},$

so

$\alpha(\beta)\succcurlyeq(\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}^{-})^{\infty}=\mathbb{L}({\mathbf{S}})^{+}s_{1}\dots
s_{l}\mathbb{L}({\mathbf{S}})^{r+1}(\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}^{-})^{\infty}\succ\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty},$

where the last inequality follows from the definition of
$\mathbb{L}({\mathbf{S}})$. This contradicts that
$\beta\in[\beta_{\ell},\beta_{*}]$.

Now there are two cases left to consider.

  1. (a)

${\mathbf{S}}^{-}$ is a proper prefix of $\mathbf{u}$. Recall that $m<j_{1}$.
Since

$u_{1}\dots u_{j_{1}-m}=s_{1}\dots
s_{j_{1}-m}\preccurlyeq\mathbb{L}({\mathbf{S}})_{m+1:j_{1}}\prec\mathbb{L}({\mathbf{S}})^{+}_{m+1:j_{1}}=\alpha_{m+1}\dots\alpha_{j_{1}},$

it follows that $\alpha_{1}\dots\alpha_{m}u_{1}\dots
u_{j_{1}-m}\prec\alpha_{1}\dots\alpha_{j_{1}}$, which yields
$\sigma^{l-m}((\mathbf{u}^{+})^{\infty})\prec\alpha(\beta)$.

  2. (b)

There is an index $i<\min\\{l,j_{1}\\}$ such that $u_{1}\dots
u_{i-1}=s_{1}\dots s_{i-1}$ and $u_{i}<s_{i}$. Then again, $u_{1}\dots
u_{i}\prec s_{1}\dots s_{i}\preccurlyeq\alpha_{m+1}\dots\alpha_{m+i}$, with
the same conclusion as in case (a).

Therefore, $\mathbf{u}^{+}\in L^{*}(\beta)$, and
$\mathbf{v}_{1}^{*}=\mathbf{u}^{+}$.

Next, let $k\geq 2$ and assume $\mathbf{v}_{k}\neq\mathbf{v}_{k-1}$. By the
same argument as above, $\mathbf{u}$ cannot be a prefix of
$\mathbf{v}_{k}^{*}$, so $\mathbf{v}_{k}^{*}=(\mathbf{u}^{+})^{*}$. We first
make a few observations:

  1. 1.

$\mathbf{u}$ cannot be a prefix of $\mathbf{v}_{k-1}$. Suppose it is. Recall
that $\mathbf{v}_{k}^{\infty}\preccurlyeq\mathbf{v}_{k-1}^{\infty}$. Since
$|\mathbf{v}_{k-1}|<j_{k}$ and $\mathbf{v}_{k}$ begins with
$\mathbf{u}\alpha_{1}\dots\alpha_{j_{k}-1}$, the constraint (10.2) implies
that $\mathbf{v}_{k-1}=\mathbf{u}\alpha_{1}\dots\alpha_{m}$ for some
$m<j_{k}$. But then $\mathbf{v}_{k-1}$ is a prefix of $\mathbf{v}_{k}$,
contradicting Lemma 10.8 (i).

  2. 2.

$\mathbf{u}$ does not extend $\mathbf{v}_{k-1}$, because $\mathbf{v}_{k}$ does
not extend $\mathbf{v}_{k-1}$.

This leaves the following cases, in which we set $l:=|\mathbf{u}|$.

  1. (a)

If $\mathbf{u}=\mathbf{v}_{k-1}^{-}$, it follows immediately that
$(\mathbf{u}^{+})^{*}=\mathbf{v}_{k-1}^{*}$.

  2. (b)

If $\mathbf{v}_{k-1}$ properly extends $\mathbf{u}^{+}$ and
$\mathbf{v}_{k-1}\in L^{*}(\beta)$, then $\mathbf{u}^{+}\in L^{*}(\beta)$ as
well by Lemma 10.2.

  3. (c)

If $\mathbf{v}_{k-1}$ properly extends $\mathbf{u}^{+}$ and $\mathbf{v}_{k-1}$
is not $\beta$-Lyndon, then for any $m$ such that $\mathbf{u}$ ends in
$\alpha_{1}\dots\alpha_{m}^{-}$ we must have $m<j_{k-1}$, since otherwise
$\mathbf{v}_{k-1}$ contains the word $\alpha_{1}\dots\alpha_{j_{k-1}}$
prematurely, contradicting Lemma 10.9. Furthermore,
$\mathbf{v}_{k-1}\prec\alpha_{m+1}\dots\alpha_{m+n_{k-1}}$ by Lemma 10.10
(ii). Since $\mathbf{u}^{+}$ and $\mathbf{v}_{k-1}$ are both Lyndon, this
implies
$(\mathbf{u}^{+})^{\infty}\prec\mathbf{v}_{k-1}0^{\infty}\prec\sigma^{m}(\alpha(\beta))$,
and hence $\sigma^{l-m}((\mathbf{u}^{+})^{\infty})\prec\alpha(\beta)$. Thus,
$\mathbf{u}^{+}\in L^{*}(\beta)$.

  4. (d)

If none of (a)-(c) above hold, then there is an index
$i<\min\\{|\mathbf{u}|,|\mathbf{v}_{k-1}|\\}$ such that $u_{1}\dots u_{i}\prec
v_{k-1,1}\dots v_{k-1,i}$. Then $u_{1}\dots
u_{i}\prec\alpha_{m+1}\dots\alpha_{m+i}$ for any $m$ by Lemma 10.10 (i). So if
$\mathbf{u}=u_{1}\dots u_{l-m}\alpha_{1}\dots\alpha_{m}^{-}$ for some $m\geq
1$, then $\sigma^{l-m}((\mathbf{u}^{+})^{\infty})\prec\alpha(\beta)$. Thus,
$\mathbf{u}^{+}\in L^{*}(\beta)$.

From these four cases and $\mathbf{v}_{k}^{*}=(\mathbf{u}^{+})^{*}$, (10.10)
follows. ∎

######  Proposition 10.13.

Let $k,\ell\in\mathbb{N}$ with $k<\ell$, and suppose
$I_{k}=[\underline{t}_{k},\overline{t}_{k})$ and
$I_{\ell}=[\underline{t}_{\ell},\overline{t}_{\ell})$ belong to $\mathcal{I}$.
Then:

  1. (i)

Either $\overline{t}_{\ell}\leq\underline{t}_{k}$ or $I_{\ell}\subseteq
I_{k}$. In other words, $I_{\ell}$ is either contained in $I_{k}$ or else lies
completely to the left of $I_{k}$.

  2. (ii)

$I_{\ell}\subseteq I_{k}$ if and only if

(10.12) $\mathbf{v}_{\ell}=\mathbf{v}_{k}\quad\mbox{or}\quad\mathbf{v}_{\ell}\
\mbox{begins with}\
\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}\
\mbox{for some $n\geq 0$}.$

###### Proof.

It suffices to show that (10.12) implies $I_{\ell}\subseteq I_{k}$, and
$\overline{t}_{\ell}>\underline{t}_{k}$ implies (10.12).

Assume first that (10.12) holds. If $\mathbf{v}_{\ell}=\mathbf{v}_{k}$, then
$\underline{t}_{\ell}>\underline{t}_{k}$ and
$\overline{t}_{\ell}=\overline{t}_{k}$, so $I_{\ell}\subseteq I_{k}$. If
$\mathbf{v}_{\ell}\neq\mathbf{v}_{k}$, then by iterating Lemma 10.8,
$\mathbf{v}_{\ell}^{\infty}\prec\mathbf{v}_{k}^{\infty}$ and
$\mathbf{v}_{\ell}$ does not extend $\mathbf{v}_{k}$. Hence, we also have
$(\mathbf{v}_{\ell}^{*})^{\infty}\preccurlyeq(\mathbf{v}_{k}^{*})^{\infty}$
since the map $\mathbf{v}\mapsto\mathbf{v}^{*}$ which sends a Lyndon word
$\mathbf{v}$ to the smallest $\beta$-Lyndon word $\mathbf{v}^{*}$ greater than
or equal to $\mathbf{v}$, is clearly nondecreasing. Therefore,
$\overline{t}_{\ell}\leq\overline{t}_{k}$. On the other hand,
$\mathbf{v}_{\ell}$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$
for some $n\geq 0$, and this implies
$\underline{t}_{\ell}\geq\underline{t}_{k}$. Thus, $I_{\ell}\subseteq I_{k}$.

Next, we show by induction that for all $\ell\geq k$,
$\overline{t}_{\ell}>\underline{t}_{k}$ implies (10.12). This is trivial for
$\ell=k$, so take $\ell^{\prime}>k$ and assume that
$\overline{t}_{\ell}>\underline{t}_{k}$ implies (10.12) for all
$k\leq\ell<\ell^{\prime}$. Suppose
$\overline{t}_{\ell^{\prime}}>\underline{t}_{k}$, so

(10.13)
$(\mathbf{v}_{\ell^{\prime}}^{*})^{\infty}\succ\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}.$

Assume $\mathbf{v}_{\ell^{\prime}}\neq\mathbf{v}_{k}$, as otherwise there is
nothing to show. We consider two cases:

Case 1. $\mathbf{v}_{\ell^{\prime}}\in L^{*}(\beta)$. Then
$\mathbf{v}_{\ell^{\prime}}^{*}=\mathbf{v}_{\ell^{\prime}}$ and hence

$\mathbf{v}_{\ell^{\prime}}^{\infty}\succ\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}.$

Since $\mathbf{v}_{k}$ and $\mathbf{v}_{\ell^{\prime}}$ are both Lyndon,
$\mathbf{v}_{\ell^{\prime}}^{\infty}$ cannot begin with $\mathbf{v}_{k}$, and
hence it must begin with $\mathbf{v}_{k}^{-}$. Thus, in view of the constraint
(10.2), there exists some integer $r\geq 0$ such that
$\mathbf{v}_{\ell^{\prime}}^{\infty}$ begins with

(10.14)
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}.$

Suppose, by way of contradiction, that $\mathbf{v}_{\ell^{\prime}}$ is a
proper prefix of this last word, so

$\mathbf{v}_{\ell^{\prime}}=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{p}\alpha_{1}\dots\alpha_{m}\qquad\mbox{for
some $0\leq p\leq r$ and $1\leq m<j_{k}$}.$

(Note $\mathbf{v}_{\ell^{\prime}}$ cannot end in
$\alpha_{1}\dots\alpha_{j_{k}}^{-}$ because $\mathbf{v}_{\ell^{\prime}}$ is
Lyndon.) Then it is easy to see that $\mathbf{v}_{\ell^{\prime}}$ does not
contain the word $\alpha_{1}\dots\alpha_{j_{k}}$ (this follows from Lemma
10.10). However, $\mathbf{v}_{\ell^{\prime}}^{\infty}$ does contain
$\alpha_{1}\dots\alpha_{j_{k}}$ because it begins with (10.14). Therefore, it
must be that
$\alpha_{1}\dots\alpha_{m}\mathbf{v}_{k}^{-}\alpha_{1}\dots\alpha_{j_{k}}^{-}$
contains $\alpha_{1}\dots\alpha_{j_{k}}$, and this word must begin inside the
block $\alpha_{1}\dots\alpha_{m}$. So let $i<m$ be such that
$\alpha_{i+1}\dots\alpha_{m}\mathbf{v}_{k}^{-}\alpha_{1}\dots\alpha_{j_{k}}^{-}$
begins with $\alpha_{1}\dots\alpha_{j_{k}}$. Then necessarily
$j_{k}-(m-i)<|\mathbf{v}_{k}|=n_{k}$ in view of Lemma 10.10. But then we have

$v_{k,1}\dots v_{k,j_{k}-(m-i)}=\alpha_{m-i+1}\dots\alpha_{j_{k}},$

so again using Lemma 10.10 it follows that

$v_{k,j_{k}-(m-i)+1}\dots
v_{k,n_{k}}\preccurlyeq\alpha_{j_{k}+1}\dots\alpha_{m-i+n_{k}}=v_{k,1}\dots
v_{k,n_{k}+m-i-j_{k}},$

contradicting that $\mathbf{v}_{k}$ is Lyndon. Therefore,
$\mathbf{v}_{\ell^{\prime}}$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$.

Case 2. $\mathbf{v}_{\ell^{\prime}}\not\in L^{*}(\beta)$. Let
$\ell\leq\ell^{\prime}$ be the largest integer such that
$\mathbf{v}_{\ell}=\mathbf{v}_{\ell^{\prime}}$. Then
$\mathbf{v}_{\ell}\neq\mathbf{v}_{\ell-1}$,
$\mathbf{v}_{\ell}\neq\mathbf{v}_{k}$ since
$\mathbf{v}_{\ell^{\prime}}\neq\mathbf{v}_{k}$, and of course
$\mathbf{v}_{\ell}\not\in L^{*}(\beta)$. By Lemma 10.12,

$\mathbf{v}_{\ell}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{\ell}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{\ell}}$

for some word $\mathbf{u}$ not ending in
$\alpha_{1}\dots\alpha_{j_{\ell}}^{-}$ and some $r\geq 0$. If
$\mathbf{u}=\mathbf{v}_{k}^{-}$, then clearly $\mathbf{v}_{\ell}$ extends
$\mathbf{v}_{k}^{-}\alpha_{1}\dots\alpha_{j_{k}}$. So assume
$\mathbf{u}\neq\mathbf{v}_{k}^{-}$. Then by Lemma 10.12 there are two
possibilities:

(a) $\mathbf{u}=\mathbf{v}_{\ell-1}^{-}$. Then
$\mathbf{v}_{\ell^{\prime}}^{*}=\mathbf{v}_{\ell}^{*}=\mathbf{v}_{\ell-1}^{*}$,
so
$(\mathbf{v}_{\ell-1}^{*})^{\infty}\succ\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}$
by (10.13). This means $\overline{t}_{\ell-1}>\underline{t}_{k}$, so by the
induction hypothesis, $\mathbf{v}_{\ell-1}$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$
for some $n\geq 0$, since
$\mathbf{v}_{\ell-1}=\mathbf{u}^{+}\neq\mathbf{v}_{k}$. Now either
$\mathbf{v}_{\ell-1}=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$,
in which case

$\mathbf{v}_{\ell}=\mathbf{v}_{\ell-1}^{-}(\alpha_{1}\dots\alpha_{j_{\ell}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{\ell}}=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n+1}(\alpha_{1}\dots\alpha_{j_{\ell}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{\ell}},$

and so $\mathbf{v}_{\ell}$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n+1}\alpha_{1}\dots\alpha_{j_{k}}$
because $j_{\ell}>j_{k}$; or $\mathbf{v}_{\ell-1}$ properly extends
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$,
in which case $\mathbf{u}$, and hence $\mathbf{v}_{\ell}$, begins with
$(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$.

(b) $\mathbf{u}\neq\mathbf{v}_{\ell-1}^{-}$. Then
$\mathbf{v}_{\ell^{\prime}}^{*}=\mathbf{v}_{\ell}^{*}=\mathbf{u}^{+}$, and
hence,

$(\mathbf{u}^{+})^{\infty}\succ\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}.$

This implies that $(\mathbf{u}^{+})^{\infty}$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$
for some $r\geq 0$. The same argument as in Case 1 then shows that
$\mathbf{u}^{+}$ must extend
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$.
If $\mathbf{u}^{+}$ properly extends this word, then so does $\mathbf{u}$ and
hence so does $\mathbf{v}_{\ell}$.

Otherwise,
$\mathbf{u}^{+}=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$,
and so

$\mathbf{v}_{\ell}=\mathbf{u}\alpha_{1}\dots\alpha_{j_{\ell}}=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r+1}\alpha_{1}\dots\alpha_{j_{\ell}},$

so that $\mathbf{v}_{\ell}$ extends
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r+1}\alpha_{1}\dots\alpha_{j_{k}}$,
because $j_{\ell}>j_{k}$.

In both case (a) and case (b) we have shown that
$\mathbf{v}_{\ell^{\prime}}=\mathbf{v}_{\ell}$ extends
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{n}\alpha_{1}\dots\alpha_{j_{k}}$
for some $n\geq 0$. This completes the proof. ∎

##  11\. Proof of Theorem 10.3

We are now ready to prove Theorem 10.3. The theorem involves three separate
statements, which we prove in Propositions 11.3, 11.5 and 11.6, respectively.
Our notation and assumptions are the same as in the previous section.

######  Lemma 11.1.

For each $k\geq 1$, $(\mathbf{v}_{k}^{*})^{\infty}\prec{\mathbf{S}}^{\infty}$.

###### Proof.

First, take $k=1$. If $\mathbf{v}_{1}\in L^{*}(\beta)$, then
$(\mathbf{v}_{1}^{*})^{\infty}=\mathbf{v}_{1}^{\infty}\prec{\mathbf{S}}^{-}1^{\infty}\prec{\mathbf{S}}^{\infty}$,
using (10.5) and the fact that ${\mathbf{S}}$ and $\mathbf{v}_{1}$ are both
Lyndon. If $\mathbf{v}_{1}\not\in L^{*}(\beta)$, then
$\mathbf{v}_{1}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{1}}$
for some word $\mathbf{u}$ and integer $r\geq 0$ by Lemma 10.12, and
$\mathbf{v}_{1}^{*}=\mathbf{u}^{+}$. This implies that
$\mathbf{v}_{1}^{*}\neq{\mathbf{S}}$, for else we would have

$\mathbf{v}_{1}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{1}}={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{r}\mathbb{L}({\mathbf{S}})^{+},$

contradicting (10.5). But
$(\mathbf{v}_{1}^{*})^{\infty}\preccurlyeq{\mathbf{S}}^{\infty}$, because
$\mathbf{v}_{1}^{\infty}\preccurlyeq{\mathbf{S}}^{\infty}$, ${\mathbf{S}}\in
L^{*}(\beta)$, and $\mathbf{v}_{1}^{*}$ is the smallest $\beta$-Lyndon word
greater than or equal to $\mathbf{v}_{1}$. Therefore,
$(\mathbf{v}_{1}^{*})^{\infty}\prec{\mathbf{S}}^{\infty}$.

For $k\geq 2$, the statement now follows inductively by Lemma 10.8. ∎

######  Definition 11.2.

We call an interval $I_{k}\in\mathcal{I}$ a non-transitivity window if it is a
maximal element of $\mathcal{I}$ with respect to set inclusion; that is, if
$I_{k}$ is not contained in any other interval $I_{\ell}\in\mathcal{I}$. We
denote by $\mathcal{I}^{\prime}$ the collection of all non-transitivity
windows.

By Proposition 10.13, the collection $\mathcal{I}^{\prime}$ is pairwise
disjoint. The name “non-transitivity window” is made clear by the next
proposition.

######  Proposition 11.3.

Let $I_{k}\in\mathcal{I}^{\prime}$ be a non-transitivity window. Then for any
$\beta$-Lyndon interval $[t_{L},t_{R}]$ with $t_{R}\in I_{k}$, the subshift
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive.

###### Proof.

Suppose $t_{R}\in I_{k}$. We first assume $I_{k}$ is open on the right, so

(11.1)
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}=b(\underline{t}_{k},\beta)\preccurlyeq
b(t_{R},\beta)\prec b(\overline{t}_{k},\beta)=(\mathbf{v}_{k}^{*})^{\infty}.$

Observe that, by definition of $\mathbf{v}_{k}^{*}$, there are no
$\beta$-Lyndon words $\mathbf{w}$ such that
$\mathbf{v}_{k}^{\infty}\prec\mathbf{w}^{\infty}\prec(\mathbf{v}_{k}^{*})^{\infty}$,
and so we actually have $b(t_{R},\beta)\prec\mathbf{v}_{k}^{\infty}$. (Strict
inequality, because if
$\mathbf{v}_{k}^{\infty}\prec(\mathbf{v}_{k}^{*})^{\infty}$, then
$\mathbf{v}_{k}\not\in L^{*}(\beta)$.) But then, since $\mathbf{v}_{k}\in
L^{*}$, $b(t_{R},\beta)$ must begin with $\mathbf{v}_{k}^{-}$.

We claim that
$\alpha_{1}\dots\alpha_{j_{k}}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$.
To see this, consider two cases:

  1. (i)

If $t_{R}>\underline{t}_{k}$, then $b(t_{R},\beta)$ begins with
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{k}}$
for some $r\geq 0$, and since
$b(t_{R},\beta)\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$, this implies that
$\alpha_{1}\dots\alpha_{j_{k}}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$.

  2. (ii)

If $t_{R}=\underline{t}_{k}$, then
$b(t_{R},\beta)=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}$,
and in this case the sequence

$\alpha_{1}\dots\alpha_{j_{k}}\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}$

lies in $\widetilde{\mathcal{K}}_{\beta}(t_{R})$, so again
$\alpha_{1}\dots\alpha_{j_{k}}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$.

From the construction of $\mathbf{v}_{k}$, (11.1) and the definition of
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$, it follows that any sequence in
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning with
$\alpha_{1}\dots\alpha_{j_{k}}$ must be of the form

(11.2)
$\alpha_{1}\dots\alpha_{j_{k}}\mathbf{v}_{k}^{p_{1}}\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{q_{1}}\,\alpha_{1}\dots\alpha_{j_{k}}\mathbf{v}_{k}^{p_{2}}\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{q_{2}}\dots,$

where $0\leq p_{i},q_{i}\leq\infty$ for all $i$. If $k>1$, then no such
sequence ends in ${\mathbf{S}}^{\infty}$, because
$\mathbf{v}_{k}^{\infty}\prec{\mathbf{S}}^{\infty}$ by Lemma 11.1 and
$\alpha_{1}\dots\alpha_{j_{k}}^{-}$ begins with
$\mathbb{L}({\mathbf{S}})^{+}$. Since ${\mathbf{S}}^{\infty}$ is always a
valid sequence in $\widetilde{\mathcal{K}}_{\beta}(t_{R})$, we see that
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive in this case.

Suppose now that $k=1$. Then we have to be a bit more careful since
$\alpha_{1}\dots\alpha_{j_{1}}^{-}=\mathbb{L}({\mathbf{S}})$. If
$t_{R}>\underline{t}_{1}$, then we cannot have $q_{i}=\infty$ for any $i$ in
(11.2), so again the legal sequences of the form (11.2) cannot end in
${\mathbf{S}}^{\infty}$, and $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not
transitive. Suppose $t_{R}=\underline{t}_{1}$, so
$b(t_{R},\beta)=\mathbf{v}_{1}^{-}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{\infty}$.
Since $b(t_{R},\beta)$ is purely periodic, this implies that either
$\mathbf{v}_{1}=\alpha_{m+1}\dots\alpha_{j_{1}}$ for some $m<j_{1}$, or else

$\mathbf{v}_{1}=\alpha_{m+1}\dots\alpha_{j_{1}}^{-}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{1}}$

for some $m<j_{1}$ and $r\geq 0$. However, it can be seen from the
inequalities in (10.6) that neither of these is possible. Thus, we must have
$t_{R}>\underline{t}_{1}$, and we are done.

Finally, we consider the case when $I_{k}$ is closed. Recall that this means
that $\mathcal{I}=\\{I_{1},\dots,I_{k}\\}$ is finite, $\mathbf{v}_{k}\in
L^{*}(\beta)$, and $\sigma^{j_{k}}(\alpha(\beta))=\mathbf{v}_{k}^{\infty}$. If
$t_{R}<\overline{t}_{k}$, the proof goes as before, so assume
$t_{R}=\overline{t}_{k}$. Then $b(t_{R},\beta)=\mathbf{v}_{k}^{\infty}$. Now
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive because
$\alpha_{1}\dots\alpha_{j_{k}}$ is a legal word that can only be followed by
the sequence $\mathbf{v}_{k}^{\infty}$. ∎

######  Remark 11.4.

In the last case in the above proof, when $I_{k}$ is closed and
$t=\overline{t}_{k}$, we showed that $\widetilde{\mathcal{K}}_{\beta}(t)$ is
not transitive. However, in this special case the set

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

is in fact itself a transitive subshift, as is easy to see. So the non-
transitivity of $\widetilde{\mathcal{K}}_{\beta}(t)$ in this case is somewhat
artificial, caused by enlarging the set $\mathcal{K}_{\beta}(t)$ more than
what is needed to obtain a subshift.

######  Proposition 11.5.

Assume ${\mathbf{S}}={\mathbf{s}}\in\Lambda_{1}=F^{*}$. Let
$t_{R}\in(0,\tau(\beta))$ be a right endpoint of a $\beta$-Lyndon interval,
and suppose $t_{R}$ does not lie in any non-transitivity window. Then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive.

(The statement of this proposition fails when ${\mathbf{S}}\in\Lambda_{k}$
with $k\geq 2$, as we will see in Section 12.)

###### Proof.

Let $\mathbf{w}$ be the $\beta$-Lyndon word such that
$b(t_{R},\beta)=\mathbf{w}^{\infty}$. We enumerate $\mathcal{I}^{\prime}$ as
$\mathcal{I}^{\prime}=\\{I_{1}=I_{k_{1}},I_{k_{2}},\dots\\}$, where
$k_{1}<k_{2}<\dots$, or as
$\mathcal{I}^{\prime}=\\{I_{1}=I_{k_{1}},I_{k_{2}},\dots,I_{k_{N}}\\}$, where
$k_{1}<k_{2}<\dots<k_{N}$, depending on whether $\mathcal{I}^{\prime}$ is
infinite or finite. Recall that the intervals $I_{k_{1}},I_{k_{2}},\dots$
proceed from right to left. There are three cases to consider:

  * •

$t_{R}\geq\overline{t}_{1}$;

  * •

$\overline{t}_{k_{i+1}}\leq t_{R}<\underline{t}_{k_{i}}$ for some $i$;

  * •

$t_{R}<\underline{t}_{k}$ for all $k$.

The argument differs somewhat for each of the three cases.

Case 1. $t_{R}\geq\overline{t}_{1}$. Then
$\mathbf{w}^{\infty}\succcurlyeq(\mathbf{v}_{1}^{*})^{\infty}\succcurlyeq\alpha_{j_{1}+1}\alpha_{j_{1}+2}\dots$.
In fact, we can show that

(11.3) $\mathbf{w}^{\infty}\succ\alpha_{j_{1}+1}\alpha_{j_{1}+2}\dots,$

by considering the following cases:

  * •

If $t_{R}>\overline{t}_{1}$, then
$\mathbf{w}^{\infty}\succ(\mathbf{v}_{1}^{*})^{\infty}$;

  * •

If $t_{R}=\overline{t}_{1}$ and $\mathbf{v}_{1}$ is not $\beta$-Lyndon, then
$(\mathbf{v}_{1}^{*})^{\infty}\succ\mathbf{v}_{1}^{\infty}\succcurlyeq\alpha_{j_{1}+1}\alpha_{j_{1}+2}\dots$
by the definition of $\mathbf{v}_{1}$;

  * •

If $t_{R}=\overline{t}_{1}$ and $\mathbf{v}_{1}$ is $\beta$-Lyndon, then,
since we assumed that $t_{R}\not\in I_{1}$, the interval $I_{1}$ was taken to
be half open and therefore
$\sigma^{j_{1}}(\alpha(\beta))\neq\mathbf{v}_{1}^{\infty}$, which implies
$(\mathbf{v}_{1}^{*})^{\infty}=\mathbf{v}_{1}^{\infty}\succ\alpha_{j_{1}+1}\alpha_{j_{1}+2}\dots$.

By (11.3), we have

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\\{\mathbf{z}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{\infty}\
\forall\,n\geq 0\\}.$

Thus, the word $\alpha_{1}\dots\alpha_{j_{1}}$ cannot occur in any sequence
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$; in particular, it
cannot occur in $\mathbf{w}^{\infty}$.

Let $\mathbf{u}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given. By Lemma 6.3,
we can extend $\mathbf{z}$ to the left to a sequence
$\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning with
$\mathbb{L}({\mathbf{s}})$. Thus, without loss of generality, we will assume
that $\mathbf{z}$ itself begins with $\mathbb{L}({\mathbf{s}})$.

If $\mathbf{u}$ does not end in $\alpha_{1}\dots\alpha_{m}$ for any $m$, set
$\mathbf{u}^{\prime}:=\mathbf{u}\mathbb{L}({\mathbf{s}})$. Otherwise, let $m$
be the largest integer such that $\mathbf{u}$ ends in
$\alpha_{1}\dots\alpha_{m}$, say $\mathbf{u}=u_{1}\dots
u_{l}\alpha_{1}\dots\alpha_{m}$. Then $m<j_{1}$, and we set
$\mathbf{u}^{\prime}:=u_{1}\dots
u_{l}\alpha_{1}\dots\alpha_{j_{1}}^{-}=u_{1}\dots
u_{l}\mathbb{L}({\mathbf{s}})$.

Observe that
$\mathbf{w}^{\infty}\prec{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}$,
so there exists $N_{1}\in\mathbb{N}$ such that
$\mathbf{w}^{\infty}\prec{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{N_{1}}0^{\infty}$.
Furthermore,
$\sigma^{n}(\mathbf{w}^{\infty})\preccurlyeq\mathbb{L}({\mathbf{s}})^{\infty}$
for all $n\geq 0$, and in fact the inequality is strict because $\mathbf{w}$
and ${\mathbf{s}}$ are both Lyndon, and $\mathbf{w}\neq{\mathbf{s}}$. Thus,
since $\mathbf{w}^{\infty}$ is periodic, there exists $N_{2}\in\mathbb{N}$
such that

$\sigma^{n}(\mathbf{w}^{\infty})\prec\mathbb{L}({\mathbf{s}})^{N_{2}}0^{\infty}\qquad\forall\,n\geq
0.$

Now let $N:=\max\\{N_{1},N_{2}\\}$, and set
$\mathbf{v}:=\mathbb{L}({\mathbf{s}})^{N}$. Then
$\mathbf{u}^{\prime}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
Therefore, $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive.

Case 2. $\overline{t}_{k_{i+1}}\leq t_{R}<\underline{t}_{k_{i}}$ for some $i$.
For ease of notation, put $k:=k_{i}$ and $\ell:=k_{i+1}$, so we have

$\overline{t}_{\ell}\leq
t_{R}<\underline{t}_{\nu}\qquad\forall\,1\leq\nu<\ell,$

since $\underline{t}_{\nu}\geq\underline{t}_{k}$ for $k\leq\nu<\ell$. By the
same reasoning as in Case 1, we have that

$\mathbf{w}^{\infty}\succ\alpha_{j_{\ell}+1}\alpha_{j_{\ell}+2}\dots,$

and so

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\\{\mathbf{z}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq(\alpha_{1}\dots\alpha_{j_{\ell}}^{-})^{\infty}\
\forall\,n\geq 0\\}.$

Thus, the word $\alpha_{1}\dots\alpha_{j_{\ell}}$ never occurs in any sequence
in $\widetilde{\mathcal{K}}_{\beta}(t_{R})$. On the other hand, for each
$1\leq\nu<\ell$,

$\mathbf{w}^{\infty}=b(t_{R},\beta)\prec
b(\underline{t}_{\nu},\beta)=\mathbf{v}_{\nu}^{-}(\alpha_{1}\dots\alpha_{j_{\nu}}^{-})^{\infty},$

so there is an integer $r_{\nu}\geq 0$ such that

(11.4)
$\mathbf{w}^{\infty}\prec\mathbf{v}_{\nu}^{-}(\alpha_{1}\dots\alpha_{j_{\nu}}^{-})^{r_{\nu}}0^{\infty}.$

Let $\mathbf{u}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given. As in Case 1,
we may assume that $\mathbf{z}$ begins with the word
$\mathbb{L}({\mathbf{s}})$. If no suffix of $\mathbf{u}$ is a prefix of
$\alpha(\beta)$, set $m=0$. Otherwise, let $m$ be the largest integer such
that $\mathbf{u}$ ends in $\alpha_{1}\dots\alpha_{m}$. In either case
$m<j_{\ell}$, and we define

$\mathbf{u}^{\prime}:=\mathbf{u}\alpha_{m+1}\dots\alpha_{j_{\ell}}^{-}(\alpha_{1}\dots\alpha_{j_{\ell-1}}^{-})^{r_{\ell-1}}\dots(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r_{1}}.$

It is not difficult to check using (11.4) that
$\mathbf{u}^{\prime}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$. (The
idea is that $\alpha_{1}\dots\alpha_{j_{\nu}}^{-}$ ends in
$\mathbf{v}_{\nu-1}^{-}$. So we append a high enough power of
$\alpha_{1}\dots\alpha_{j_{\nu-1}}^{-}$ to it to ensure that
$\sigma^{n}(\mathbf{u}^{\prime}\mathbf{z})\succcurlyeq\mathbf{w}^{\infty}$ for
all $n$, and repeat until we reach $\nu=1$.) Hence,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive.

Case 3. $t_{R}<\underline{t}_{k}$ for all $k$. This case is the most involved.

(a) Assume first that $\mathcal{I}$ is infinite. Then as in Case 2 there exist
integers $r_{1},r_{2},\dots$ such that

$\mathbf{w}^{\infty}\prec\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r_{k}}0^{\infty},\qquad
k=1,2,\dots.$

Let $\mathbf{u}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given. If
$\mathbf{u}$ does not end in $\alpha_{1}\dots\alpha_{m}$ for any $m$, set
$\mathbf{u}^{\prime}:=\mathbf{u}$. Otherwise, let $m$ be the largest integer
such that $\mathbf{u}$ ends in $\alpha_{1}\dots\alpha_{m}$. Here we do not
have an upper bound for $m$, however there is an index $k$ such that
$j_{k-1}\leq m<j_{k}$, since $j_{k}\to\infty$ as $k\to\infty$. We can thus
extend $\mathbf{u}$ to a word
$\mathbf{u}^{\prime}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$
ending in the block

$\alpha_{1}\dots\alpha_{j_{k}}^{-}(\alpha_{1}\dots\alpha_{j_{k-1}}^{-})^{r_{k-1}}\dots(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r_{1}}.$

Similarly, as before, we may assume $\mathbf{z}$ begins with
$\mathbb{L}({\mathbf{s}})$. Then
$\mathbf{u}^{\prime}\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$
just in Case 2.

(b) Assume next that $\mathcal{I}$ is finite:
$\mathcal{I}=\\{I_{1},I_{2},\dots,I_{k_{0}}\\}$, where $k_{0}\in\mathbb{N}$.
This case is more difficult because $\mathbf{u}$ can end in an arbitrarily
long prefix of $\alpha(\beta)$, but there are only finitely many $j_{k}$’s, so
the construction of case (a) above needs to be modified. Set
$n_{0}:=|\mathbb{L}({\mathbf{s}})^{+}\mathbf{v}_{1}\mathbf{v}_{2}\dots\mathbf{v}_{k_{0}}|=j_{k_{0}}+|\mathbf{v}_{k_{0}}|$.

(i) Suppose first that $\alpha(\beta)$ is not periodic. Note that
$\sigma^{n_{0}}(\alpha(\beta))\preccurlyeq\mathbf{v}_{k_{0}}^{\infty}$, so
$\sigma^{n_{0}}(\alpha(\beta))$ begins with $0$. Since
$\sigma^{n_{0}}(\alpha(\beta))\neq 0^{\infty}$, there is an index
$n_{1}>n_{0}$ such that $\alpha_{n_{0}+1}\dots\alpha_{n_{1}}$ is Lyndon. Then
$\sigma^{n_{1}}(\alpha(\beta))\succ(\alpha_{n_{0}+1}\dots\alpha_{n_{1}})^{\infty}$,
or else the word $\alpha_{n_{0}+1}\dots\alpha_{n_{1}}$ would have become
$\mathbf{v}_{k_{0}+1}$ and generated an additional interval $I_{k_{0}+1}$ in
$\mathcal{I}$. Hence, there is an index $n_{2}>n_{1}$ such that
$\alpha_{n_{0}+1}\dots\alpha_{n_{2}}$ is also Lyndon. Repeating this argument
yields an infinite, strictly increasing sequence $(n_{i})_{n\in\mathbb{N}}$
such that $\alpha_{n_{0}+1}\dots\alpha_{n_{i}}$ is Lyndon for each $i$, and

$\sigma^{n_{i}}(\alpha(\beta))\succ(\alpha_{n_{0}+1}\dots\alpha_{n_{i}})^{\infty}\qquad\forall\,i\in\mathbb{N}.$

Note also that the tail $\sigma^{n_{0}}(\alpha(\beta))$ cannot be periodic. If
now $\mathbf{w}^{\infty}\succ\sigma^{n_{0}}(\alpha(\beta))$, then

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\\{\mathbf{z}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq(\alpha_{1}\dots\alpha_{n_{0}}^{-})^{\infty}\
\forall\,n\geq 0\\},$

and the proof proceeds essentially as in Case 2. Otherwise, since
$\sigma^{n_{0}}(\alpha(\beta))$ is not periodic, there is a sufficiently large
index $i_{0}$ such that

(11.5)
$\mathbf{w}^{\infty}\prec\alpha_{n_{0}+1}\dots\alpha_{n_{i_{0}}}^{-}0^{\infty}.$

Claim: The tail $\sigma^{n_{0}}(\alpha(\beta))$ does not contain the word
$\alpha_{1}\dots\alpha_{n_{0}}$.

To see this, write $(c_{i}):=\sigma^{n_{0}}(\alpha(\beta))$ and suppose by way
of contradiction that $c_{m+1}\dots
c_{m+n_{0}}=\alpha_{1}\dots\alpha_{n_{0}}$, where we can choose $m$ to be
minimal, i.e. $c_{l+1}\dots c_{l+n_{0}}\neq\alpha_{1}\dots\alpha_{n_{0}}$ for
all $0\leq l<m$. We first show that $c_{1}\dots c_{m+n_{0}}$ is Lyndon. This
means verifying that

(11.6) $c_{l+1}\dots c_{m+n_{0}}\succ c_{1}\dots
c_{m+n_{0}-l}\qquad\forall\,1\leq l<m+n_{0}.$

Recall that

$c_{m+1}\dots
c_{m+n_{0}}=\alpha_{1}\dots\alpha_{n_{0}}=\mathbb{L}({\mathbf{s}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}.$

By the same argument as in the proof of Lemma 10.8, $c_{l+1}\dots c_{m+n_{0}}$
does not extend $\mathbf{v}_{k}$ for any $k\leq k_{0}$. Since in addition each
$\mathbf{v}_{k}$ is Lyndon, this implies (11.6) for $m+j_{1}\leq l<m+n_{0}$.

Furthermore, for $m\leq l<m+j_{1}$ we have

$\displaystyle c_{l+1}\dots c_{m+j_{1}}$
$\displaystyle=\alpha_{l-m+1}\dots\alpha_{j_{1}}\succ s_{1}\dots
s_{j_{1}-(l-m)}$
$\displaystyle\succcurlyeq(\mathbf{v}_{k_{0}}^{\infty})_{1:j_{1}-(l-m)}\succcurlyeq
c_{1}\dots c_{j_{1}-(l-m)},$

where the first inequality follows since $\alpha_{l-m+1}\dots\alpha_{j_{1}}$
is a suffix of $\mathbb{L}({\mathbf{s}})^{+}$, the second inequality follows
by Lemma 11.1, and the last inequality follows from the definition of
$\mathbf{v}_{k_{0}}$.

(Note: If the collection $\mathcal{I}$ is empty, then $n_{0}=j_{1}$ and we
obtain directly the inequality $s_{1}\dots s_{j_{1}-(l-m)}\succcurlyeq
c_{1}\dots c_{j_{1}-(l-m)}$ because
$\sigma^{j_{1}}(\alpha(\beta))\prec{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}$.)

Finally, for $1\leq l<m$ we have $c_{l+1}\dots c_{m}\succcurlyeq c_{1}\dots
c_{m-l}$ because $c_{1}\dots c_{m-l}$ is a prefix of the Lyndon word
$\alpha_{n_{0}+1}\dots\alpha_{n_{i}}$ for a sufficiently large $i$; and
$c_{m+1}\dots c_{m+n_{0}}=\alpha_{1}\dots\alpha_{n_{0}}\succ c_{m-l+1}\dots
c_{m-l+n_{0}}$ by the minimality of $m$. This gives (11.6) for $1\leq l<m$ as
well.

Having verified (11.6), we conclude that $c_{1}\dots c_{m+n_{0}}$ is Lyndon.
But then $\sigma^{2n_{0}+m}(\alpha(\beta))\succ(c_{1}\dots
c_{n_{0}+m})^{\infty}$, which implies
$\sigma^{n_{0}+m}(\alpha(\beta))\succ\alpha(\beta)$, a contradiction.
Therefore, our assumption that $c_{m+1}\dots
c_{m+n_{0}}=\alpha_{1}\dots\alpha_{n_{0}}$ was invalid, and the claim is
proven.

From the Claim, it follows that

(11.7)
$\sigma^{n}(\alpha(\beta))\preccurlyeq(\alpha_{1}\dots\alpha_{n_{0}}^{-})^{\infty}\qquad\forall\,n\geq
n_{0}.$

Now let $\mathbf{u}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given. As in Case 2,
we can extend $\mathbf{z}$ to a sequence
$\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning with
$\mathbb{L}({\mathbf{s}})$. We extend $\mathbf{u}$ first to a sequence
$\mathbf{u}^{\prime}$ ending in $\alpha_{1}\dots\alpha_{n_{i_{0}}}^{-}$, where
$i_{0}$ was chosen to satisfy (11.5), in the usual way by pasting $\mathbf{u}$
and $\alpha_{1}\dots\alpha_{n_{i_{0}}}^{-}$ together along their longest
overlap. If $\mathbf{u}^{\prime}$ does not end in a prefix of $\mathbf{w}$,
then
$\mathbf{u}^{\prime}\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$
and we are done.

Otherwise, let $m$ be the largest integer such that $\mathbf{u}^{\prime}$ ends
in $w_{1}\dots w_{m}$, where we write $(w_{i}):=\mathbf{w}^{\infty}$. Then by
(11.5) and the Lyndon property of $\alpha_{n_{0}+1}\dots\alpha_{n_{i_{0}}}$,

$w_{1}\dots
w_{m}\preccurlyeq\alpha_{n_{0}+1}\dots\alpha_{n_{0}+m}\preccurlyeq\alpha_{n_{i_{0}}-m+1}\dots\alpha_{n_{i_{0}}}^{-}=w_{1}\dots
w_{m},$

hence $w_{1}\dots w_{m}=\alpha_{n_{0}+1}\dots\alpha_{n_{0}+m}$. By (11.5) and
(11.7), it follows that

$w_{m+1}w_{m+2}\dots\prec\alpha_{n_{0}+m+1}\dots\alpha_{n_{i_{0}}}^{-}0^{\infty}\prec(\alpha_{1}\dots\alpha_{n_{0}}^{-})^{\infty},$

so there is some integer $r_{*}\geq 1$ such that
$w_{m+1}w_{m+2}\dots\prec(\alpha_{1}\dots\alpha_{n_{0}}^{-})^{r_{*}}0^{\infty}$.
Recall again that
$\alpha_{1}\dots\alpha_{n_{0}}^{-}=\mathbb{L}({\mathbf{s}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-}$.
As in Case 2 there exist integers $r_{1},\dots,r_{k_{0}}$ such that

(11.8)
$\mathbf{w}^{\infty}\prec\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{r_{k}}0^{\infty},\qquad
k=1,\dots,k_{0}.$

We now extend $\mathbf{u}^{\prime}$ further by setting

$\mathbf{u}^{\prime\prime}:=\mathbf{u}^{\prime}(\alpha_{1}\dots\alpha_{n_{0}}^{-})^{r_{*}}(\alpha_{1}\dots\alpha_{j_{k_{0}}}^{-})^{r_{k_{0}}}\dots(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r_{1}}.$

Then
$\mathbf{u}^{\prime\prime}\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.
Hence, $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive.

(ii) Suppose next that $\alpha(\beta)$ is periodic. Then

$\alpha(\beta)=(\alpha_{1}\dots\alpha_{n_{0}})^{\infty}=(\mathbb{L}({\mathbf{s}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-})^{\infty}.$

Let $\mathbf{u}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta}(t_{R}))$. If
$\mathbf{u}$ does not end in a prefix of $\alpha(\beta)$, we set
$\mathbf{u}^{\prime}:=\mathbf{u}\alpha_{1}\dots\alpha_{n_{0}}$. Otherwise,
there is some $r\geq 0$ and $0\leq m<n_{0}$ such that $\mathbf{u}$ ends in
$(\alpha_{1}\dots\alpha_{n_{0}})^{r}\alpha_{1}\dots\alpha_{m}$. In that case
we extend $\mathbf{u}$ slightly further to a word $\mathbf{u}^{\prime}$ ending
in $(\alpha_{1}\dots\alpha_{n_{0}})^{r+1}$. Observe that $\mathbf{u}^{\prime}$
then ends in
$\mathbb{L}({\mathbf{s}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-}$. From
here we proceed as above: We choose exponents $r_{1},\dots,r_{k_{0}}$
satisfying (11.8), and set

$\mathbf{u}^{\prime\prime}:=\mathbf{u}^{\prime}(\alpha_{1}\dots\alpha_{j_{k_{0}}}^{-})^{r_{k_{0}}}\dots(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r_{1}}.$

This word $\mathbf{u}^{\prime\prime}$ can then be connected directly to any
sequence $\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$
beginning with $\mathbb{L}({\mathbf{s}})$. ∎

######  Proposition 11.6.

Let $I_{k}=[\underline{t}_{k},\overline{t}_{k})\in\mathcal{I}^{\prime}$ be a
non-transitivity window. Then $h(\widetilde{\mathcal{K}}_{\beta}(t))$ is
constant on $I_{k}$.

###### Proof.

This will follow from Lemmas 14.4 and 14.5 in Section 14. ∎

###### Proof of Theorem 10.3.

The theorem follows from Propositions 11.3, 11.5 and 11.6. ∎

The work we did in this section and the last, also has an important
consequence for the bifurcation sets $\mathscr{E}_{\beta}$ and
$\mathscr{B}_{\beta}$:

######  Theorem 11.7.

Let $\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$, where
${\mathbf{S}}\in\Lambda$. If $I_{k}$ is a non-transitivity window, then
$\dim_{H}(\mathscr{E}_{\beta}\cap I_{k})>0$. In particular, if the collection
$\mathcal{I}$ is nonempty, then
$\dim_{H}(\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta})>0$.

###### Proof.

Let $I_{k}=[\underline{t}_{k},\overline{t}_{k})$ be a non-transitivity window,
so

$b(\underline{t}_{k},\beta)=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty},\qquad
b(\overline{t}_{k},\beta)=(\mathbf{v}_{k}^{*})^{\infty}.$

Writing $B_{k}:=\alpha_{1}\dots\alpha_{j_{k}}$ for brevity, consider points
$t$ with greedy expansion

$b(t,\beta)=\mathbf{v}_{k}^{-}(B_{k}^{-})^{r_{1}}B_{k}\mathbf{v}_{k}^{-}(B_{k}^{-})^{r_{2}}B_{k}\mathbf{v}_{k}^{-}(B_{k}^{-})^{r_{3}}B_{k}\dots.$

Note that this expression is indeed a greedy $\beta$-expansion since

$B_{k}\mathbf{v}_{k}^{-}\prec
B_{k}\mathbf{v}_{k}=\alpha_{1}\dots\alpha_{j_{k}}\alpha_{j_{k}+1}\dots\alpha_{j_{k}+n_{k}}.$

Furthermore, observe that $\sigma^{n}(b(t,\beta))\succcurlyeq b(t,\beta)$ for
all $n\geq 0$ if and only if
$(r_{1},r_{2},\dots)\succcurlyeq(r_{n},r_{n+1},\dots)$ for all $n\geq 1$.
Taking $r_{1}=3$ and $r_{i}\in\\{1,2\\}$ arbitrarily for $i\geq 2$ we see that
there are uncountably many points of $\mathscr{E}_{\beta}$ in $I_{k}$. In
fact, the set

$C:=\\{\mathbf{v}_{k}^{-}(B_{k}^{-})^{3}B_{k}\mathbf{v}_{k}^{-}(B_{k}^{-})^{r_{2}}B_{k}\mathbf{v}_{k}^{-}(B_{k}^{-})^{r_{3}}B_{k}\dots:r_{i}\in\\{1,2\\}\
\forall\,i\geq 2\\}$

has positive entropy, hence positive Hausdorff dimension in $\Sigma_{\beta}$.
Therefore, by (4.1),

$\dim_{H}(\mathscr{E}_{\beta}\cap I_{k})\geq\dim_{H}\pi_{\beta}(C)>0.$

The second statement follows from the first and Proposition 11.6, which
implies that there are no points of $\mathscr{B}_{\beta}$ in $I_{k}$. ∎

##  12\. Higher order basic intervals

In this section we generalize Theorem 10.3 to the higher order basic interval
$[\beta_{\ell}^{{\mathbf{S}}},\beta_{r}^{{\mathbf{S}}}]$, where
${\mathbf{S}}\in\Lambda$. We do this by combining the construction from
Section 10 with the renormalization method of Section 7.

Recall that $\mathcal{T}_{R}(\beta)$ is the set of all right endpoints of
$\beta$-Lyndon intervals in $[0,\tau(\beta)]$. The following theorem
established the final case of Theorem 4.4.

######  Theorem 12.1.

Let $[\beta_{\ell},\beta_{*}]$ be a basic interval generated by a word
${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{n}\in\Lambda$,
where ${\mathbf{r}}_{i}\in F^{*}$ for each $i$, and let
$\beta\in(\beta_{\ell},\beta_{*})$. Then there is a (finite or infinite,
possibly empty) collection $\mathcal{I}$ of intervals such that

  1. (i)

For any $t_{R}\in\mathcal{T}_{R}(\beta)$,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive if and only if
$b(t_{R},\beta)\prec{\mathbf{r}}_{1}^{-}\mathbb{L}({\mathbf{r}}_{1})^{\infty}$
and $t_{R}\not\in\bigcup_{I\in\mathcal{I}}I$;

  2. (ii)

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 $b(t_{R},\beta)$;

  3. (iii)

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;

  4. (iv)

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

We begin by proving an extension of Lemma 6.3.

######  Lemma 12.2.

Let ${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{n}$, where
${\mathbf{r}}_{i}\in F^{*}$ for each $i$, and let
$\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}}]$. Let
$[t_{L},t_{R}]$ be a $\beta$-Lyndon interval such that
$b(t_{R},\beta)\prec{\mathbf{r}}_{1}^{-}\mathbb{L}({\mathbf{r}}_{1})^{\infty}$.
If $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$, then $\mathbf{z}$
can be extended to the left to a sequence
$\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning with
${\mathbf{r}}_{1}$, and also to a sequence
$\mathbf{z}^{\prime\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning
with $\mathbb{L}({\mathbf{r}}_{1})$.

The proof goes exactly like that of Lemma 6.3, replacing ${\mathbf{s}}$ with
${\mathbf{r}}_{1}$ and observing that $\alpha(\beta)$ begins with
$\mathbb{L}({\mathbf{S}})^{+}$, which in turn begins with
$\mathbb{L}({\mathbf{r}}_{1})^{+}$, followed by either ${\mathbf{r}}_{1}$ or
${\mathbf{r}}_{1}^{-}$.

Now let
${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{n}\in\Lambda$
with $n\geq 2$, and consider
$\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$. We follow
the same construction as in Section 10. Thus, $j_{1}=|{\mathbf{S}}|$, the
first special word $\mathbf{v}_{1}$ begins with $\alpha_{j_{1}+1}$, etc. The
procedure from Section 10 generates a (finite or infinite) collection
$\mathcal{I}=\\{I_{1},I_{2},\dots\\}$ of intervals.

Recall that the transitivity result of Proposition 11.5 required that
${\mathbf{S}}\in F^{*}$. For general ${\mathbf{S}}\in\Lambda$, we have instead
the following fact:

######  Proposition 12.3.

With $\beta$ as above, let $[t_{L},t_{R}]$ be a $\beta$-Lyndon interval. Then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive if and only if
$b(t_{R},\beta)\prec{\mathbf{r}}_{1}^{-}\mathbb{L}({\mathbf{r}}_{1})^{\infty}$
and $t_{R}\not\in\bigcup_{I\in\mathcal{I}}I$.

###### Proof.

Write ${\mathbf{r}}:={\mathbf{r}}_{1}$ for brevity. If
$t_{R}\in\bigcup_{I\in\mathcal{I}}I$, then
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive by Proposition
11.3. We now show that $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not
transitive when
$b(t_{R},\beta)\succcurlyeq{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
This follows essentially the same way as in the proof of Proposition 7.6. Let
$\mathbf{w}$ be the $\beta$-Lyndon word such that
$b(t_{R},\beta)=\mathbf{w}^{\infty}$. Observe that for any sequence
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$, we have

$\Phi_{\mathbf{r}}(0^{\infty})={\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}\preccurlyeq\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\preccurlyeq\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}\preccurlyeq\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{\infty}=\Phi_{\mathbf{r}}(1^{\infty}).$

Thus, if $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ begins with
${\mathbf{r}}^{-}$ or $\mathbb{L}({\mathbf{r}})^{+}$, then
$\mathbf{z}=\Phi_{\mathbf{r}}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}$ by Lemma 7.2. Furthermore, since
$\mathbf{w}^{\infty}\prec{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}$
and ${\mathbf{S}}^{-}$ begins with the word ${\mathbf{r}}^{-}$,
$\mathbf{w}^{\infty}$ also begins with ${\mathbf{r}}^{-}$. The rest of the
proof follows that of Proposition 7.6.

Next, we show that $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive when
$b(t_{R},\beta)\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$ and
$t_{R}\not\in\bigcup_{I\in\mathcal{I}}I$. This proceeds almost exactly as in
the proof of Proposition 11.5, with one essential difference: Since
$\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$,
we can find an integer $M$ such that
$\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{M}0^{\infty}$.
In each of the three cases in the proof of Proposition 11.5, we can assume
that $\mathbf{z}$ begins with $\mathbb{L}({\mathbf{r}})$ by Lemma 12.2. We now
extend $\mathbf{z}$ further by preceding it with the factor
$\mathbb{L}({\mathbf{r}})^{M}$, calling the new sequence again $\mathbf{z}$.
The rest of the proof is now the same as before (changing ${\mathbf{s}}$ to
${\mathbf{S}}$ throughout). ∎

Recall that
$X({\mathbf{s}})=\Phi_{\mathbf{s}}\big{(}\\{0,1\\}^{\mathbb{N}}\big{)}$ and
$X^{*}({\mathbf{s}})=\Phi_{\mathbf{s}}\big{(}\\{0,1\\}^{*}\big{)}$ for any
Lyndon word ${\mathbf{s}}$.

######  Lemma 12.4.

Suppose
$(\mathbf{v}_{k}^{*})^{\infty}\succ{\mathbf{r}}_{1}^{-}\mathbb{L}({\mathbf{r}}_{1})^{\infty}$.
Then $\mathbf{v}_{k}\in X^{*}({\mathbf{r}}_{1})$, and
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}\succcurlyeq{\mathbf{r}}_{1}^{-}\mathbb{L}({\mathbf{r}}_{1})^{\infty}$.

###### Proof.

We prove this by induction on $k$. Take first $k=1$, and assume
$(\mathbf{v}_{1}^{*})^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$,
where we write ${\mathbf{r}}:={\mathbf{r}}_{1}$. If $\mathbf{v}_{1}\in
L^{*}(\beta)$, then
$\mathbf{v}_{1}^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$,
so $\mathbf{v}_{1}^{\infty}$ begins with ${\mathbf{r}}^{-}$, and then
$\mathbf{v}_{1}^{\infty}\in X({\mathbf{r}})$ by Lemma 7.2. But then
$\mathbf{v}_{1}\in X^{*}({\mathbf{r}})$ because $\mathbf{v}_{1}$ is Lyndon.
Moreover, since $\mathbf{v}_{1}$ ends in 1, the last block of $\mathbf{v}_{1}$
is ${\mathbf{r}}$ or $\mathbb{L}({\mathbf{r}})^{+}$, and so
$\mathbf{v}_{1}^{-}\in X^{*}({\mathbf{r}})$ as well.

If $\mathbf{v}_{1}\not\in L^{*}(\beta)$, then
$\mathbf{v}_{1}=\mathbf{u}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{r}\alpha_{1}\dots\alpha_{j_{1}}$
for some word $\mathbf{u}$ and integer $r\geq 0$, and
$\mathbf{v}_{1}^{*}=\mathbf{u}^{+}$ by Lemma 10.12. Here $\mathbf{u}^{+}\in
L^{*}(\beta)$ and
$(\mathbf{u}^{+})^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$,
so $\mathbf{u}\in X^{*}({\mathbf{r}})$ by the same argument as above.
Observing that

$\alpha_{1}\dots\alpha_{j_{1}}^{-}=\mathbb{L}({\mathbf{S}})=\Phi_{\mathbf{r}}(\mathbb{L}({\mathbf{r}}_{2}\bullet\dots\bullet{\mathbf{r}}_{n}))\in
X^{*}({\mathbf{r}}),$

it follows that $\mathbf{v}_{1}\in X^{*}({\mathbf{r}})$ as well, and then also
$\mathbf{v}_{1}^{-}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{\infty}\in
X^{*}({\mathbf{r}})$. This implies
$\mathbf{v}_{1}^{-}(\alpha_{1}\dots\alpha_{j_{1}}^{-})^{\infty}\succcurlyeq{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.

Next, let $k\geq 2$ and assume the lemma holds for the words
$\mathbf{v}_{1},\dots,\mathbf{v}_{k-1}$. Suppose
$(\mathbf{v}_{k}^{*})^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
The same argument we used for $\mathbf{v}_{1}$ shows that $\mathbf{v}_{k}\in
X^{*}({\mathbf{r}})$. Now since
$\mathbf{v}_{l}^{\infty}\succ\mathbf{v}_{k}^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$
for all $l<k$, the induction hypothesis implies that $\mathbf{v}_{l}\in
X^{*}({\mathbf{r}})$ for all $l<k$, and thus,
$\alpha_{1}\dots\alpha_{j_{k}}=\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k-1}\in
X^{*}({\mathbf{r}})$. It follows that
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}\in
X^{*}({\mathbf{r}})$, and hence,
$\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty}\succcurlyeq{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
∎

###### Proof of Theorem 12.1.

By Theorem 10.3, the theorem holds for the case $n=1$, i.e.
${\mathbf{S}}={\mathbf{r}}\in F^{*}$; we simply set
$\mathcal{K}_{\beta}^{\prime}(t_{R}):=\widetilde{\mathcal{K}}_{\beta}(t_{R})$
for $t_{R}\in\mathcal{T}_{R}(\beta)\backslash\bigcup_{I\in\mathcal{I}}I$.

Now let ${\mathbf{S}}={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{m}$,
where $m\geq 2$ and ${\mathbf{r}}_{i}\in F^{*}$ for each $i$. Let
$\beta\in(\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$, and
construct the Lyndon words $\mathbf{v}_{1},\mathbf{v}_{2},\dots$ and
collection of intervals $\mathcal{I}=\\{I_{1},I_{2},\dots\\}$ as outlined
above. Then statement (i) follows from Proposition 12.3.

We define a new base
$\beta^{\prime}\in[\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$ as
follows. Set ${\mathbf{r}}:={\mathbf{r}}_{1}$. If
$\sigma^{n}(\alpha(\beta))\succcurlyeq{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$
for all $n\geq j_{1}$ (and hence for all $n\geq 0$), we set
$\beta^{\prime}:=\beta$ and $n_{0}:=\infty$. Otherwise, let

$n_{0}:=\min\\{n\geq
j_{1}:\sigma^{n}(\alpha(\beta))\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}\\}.$

Since the words $\mathbf{v}_{1},\mathbf{v}_{2},\dots$ are Lyndon, there is an
integer $k_{0}\geq 0$ such that $n_{0}=j_{k_{0}+1}$. We let $\beta^{\prime}$
be the base given by

$\alpha(\beta^{\prime})=(\alpha_{1}\dots\alpha_{j_{k_{0}+1}}^{-})^{\infty}=\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-}\big{)}^{\infty},$

where we interpret the last expression as
$\alpha(\beta^{\prime})=\mathbb{L}({\mathbf{S}})^{\infty}$ if $k_{0}=0$. Let
$t_{*}$ be the point given by
$b(t_{*},\beta)={\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$. By Lemma
12.4, any interval $I_{k}\in\mathcal{I}$ lies either completely in $(0,t_{*})$
or in $[t_{*},\tau(\beta)]$. By definition of $n_{0}$, the intervals
$I_{1},\dots,I_{k_{0}}$ lie to the right of the critical point $t_{*}$, and
$I_{k_{0}+1},I_{k_{0}+2},\dots$ lie to the left of $t_{*}$. (If
$n_{0}=\infty$, then each $I_{k}$ lies to the right of $t_{*}$; in this case,
we set $k_{0}:=\infty$ for convenience.)

Due to the special way in which we constructed the intervals
$I_{1},I_{2},\dots$ for the case of periodic $\alpha(\beta)$ in Section 10,
the non-transitivity windows for $\beta^{\prime}$ are exactly the intervals
$I_{1}^{\prime}=[\underline{t}_{1}^{\prime},\overline{t}_{1}^{\prime}),\dots,I_{k_{0}}^{\prime}=[\underline{t}_{k_{0}}^{\prime},\overline{t}_{k_{0}}^{\prime})$
given by

$b(\underline{t}_{k}^{\prime},\beta^{\prime})=\mathbf{v}_{k}^{-}(\alpha_{1}\dots\alpha_{j_{k}}^{-})^{\infty},\qquad
b(\overline{t}_{k}^{\prime},\beta^{\prime})=(\mathbf{v}_{k}^{*})^{\infty},\qquad
k=1,2,\dots,k_{0}.$

Let $\mathbf{w}$ be a $\beta$-Lyndon word such that
$\mathbf{w}^{\infty}\succ{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$
and $\pi_{\beta}(\mathbf{w}^{\infty})\not\in\bigcup_{k=1}^{k_{0}}I_{k}$, and
let $t_{R}$ be the point such that $b(t_{R},\beta)=\mathbf{w}^{\infty}$. We
first claim that $\mathbf{w}$ is also $\beta^{\prime}$-Lyndon. This is obvious
if $\beta^{\prime}=\beta$, so assume $\beta^{\prime}\neq\beta$ so that
$n_{0}<\infty$. Suppose
$\sigma^{l}(\mathbf{w}^{\infty})\succcurlyeq\alpha(\beta^{\prime})=\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-}\big{)}^{\infty}$
for some $l$. Since $\mathbf{w}$ is Lyndon and the words
$\mathbf{v}_{1},\dots,\mathbf{v}_{k_{0}}$ decrease lexicograpically, this can
only happen if
$\mathbf{w}=\mathbf{v}_{k_{0}}^{-}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}-1}=\mathbf{v}_{k_{0}}^{-}\alpha_{1}\dots\alpha_{j_{k_{0}}}$.
But then

$b(\underline{t}_{k_{0}},\beta)=\mathbf{v}_{k_{0}}^{-}(\alpha_{1}\dots\alpha_{j_{k_{0}}}^{-})^{\infty}\prec\mathbf{w}^{\infty}\prec(\mathbf{v}_{k_{0}}^{*})^{\infty}=b(\overline{t}_{k_{0}},\beta),$

so $t_{R}=\pi_{\beta}(\mathbf{w}^{\infty})\in I_{k_{0}}$, a contradiction.
Hence, $\mathbf{w}$ is $\beta^{\prime}$-Lyndon, and there is a point
$t_{R}^{\prime}$ such that
$b(t_{R}^{\prime},\beta^{\prime})=\mathbf{w}^{\infty}$.

Furthermore, $\sigma^{n_{0}}(\alpha(\beta))\prec\mathbf{w}^{\infty}$ if
$n_{0}$ is finite, and then

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq(\alpha_{1}\dots\alpha_{j_{k_{0}}}^{-})^{\infty}\
\forall\,n\geq 0\\}.$

It follows that
$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\widetilde{\mathcal{K}}_{\beta^{\prime}}(t_{R}^{\prime})$.

Now define
${\mathbf{S}}_{i}:={\mathbf{r}}_{1}\bullet\dots\bullet{\mathbf{r}}_{i}$ for
$i=1,2,\dots,m$, so in particular, ${\mathbf{S}}_{m}={\mathbf{S}}$. Since
$\tau(\beta)={\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty}={\mathbf{S}}_{m}^{-}\mathbb{L}({\mathbf{S}}_{m})^{\infty}$,
there is a unique $i\in\\{1,2,\dots,m-1\\}$ such that

${\mathbf{S}}_{i}^{-}\mathbb{L}({\mathbf{S}}_{i})^{\infty}\prec\mathbf{w}^{\infty}\prec{\mathbf{S}}_{i+1}^{-}\mathbb{L}({\mathbf{S}}_{i+1})^{\infty}.$

Set $\mathbf{R}:={\mathbf{S}}_{i}$ and
$\mathbf{R}^{\prime}:={\mathbf{r}}_{i+1}\bullet\dots\bullet{\mathbf{r}}_{m}$.
It follows (by Lemma 7.2 if $n_{0}=\infty$; by Lemma 12.4 if $n_{0}<\infty$)
that $\alpha(\beta^{\prime})\in X(\mathbf{R})$, and hence
$\alpha(\beta^{\prime})=\Phi_{\mathbf{R}}(\alpha(\hat{\beta}))$ for some base
$\hat{\beta}$. Note that ${\mathbf{S}}=\mathbf{R}\bullet\mathbf{R}^{\prime}$.
Since
$\beta^{\prime}\in[\beta_{\ell}^{{\mathbf{S}}},\beta_{*}^{{\mathbf{S}}})$, we
have
$\hat{\beta}\in[\beta_{\ell}^{\mathbf{R}^{\prime}},\beta_{*}^{\mathbf{R}^{\prime}})$.
We make the following observations.

  * •

If $n_{0}=\infty$ and $\mathcal{I}$ is infinite, then
$\alpha(\beta^{\prime})=\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\mathbf{v}_{2}\mathbf{v}_{3}\dots$
and
$\alpha(\hat{\beta})=\mathbb{L}(\mathbf{R}^{\prime})^{+}\hat{\mathbf{v}}_{1}\hat{\mathbf{v}}_{2}\hat{\mathbf{v}}_{3}\dots$,
where for each $k$, $\hat{\mathbf{v}}_{k}$ is a Lyndon word such that
$\Phi_{\mathbf{R}}(\hat{\mathbf{v}}_{k})=\mathbf{v}_{k}$.

  * •

If $n_{0}=\infty$ and $\mathcal{I}$ is finite, then for some $k\geq 0$,
$\alpha(\beta^{\prime})=\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k}\mathbf{y}$
and
$\alpha(\hat{\beta})=\mathbb{L}(\mathbf{R}^{\prime})^{+}\hat{\mathbf{v}}_{1}\dots\hat{\mathbf{v}}_{k}\hat{\mathbf{y}}$,
where $\mathbf{y}$ and $\hat{\mathbf{y}}$ are sequences satisfying
$\sigma^{j}(\mathbf{y})\succ\mathbf{y}$ for all $j\geq 0$,
$\mathbf{y}=\Phi_{\mathbf{R}}(\hat{\mathbf{y}})$, and hence also
$\sigma^{j}(\hat{\mathbf{y}})\succ\hat{\mathbf{y}}$ for all $j\geq 0$.

  * •

If $n_{0}<\infty$, then
$\alpha(\beta^{\prime})=\big{(}\mathbb{L}({\mathbf{S}})^{+}\mathbf{v}_{1}\dots\mathbf{v}_{k_{0}}^{-}\big{)}^{\infty}$
and
$\alpha(\hat{\beta})=\big{(}\mathbb{L}(\mathbf{R}^{\prime})^{+}\hat{\mathbf{v}}_{1}\dots\hat{\mathbf{v}}_{k_{0}}^{-}\big{)}^{\infty}$,
where again, $\mathbf{v}_{k}=\Phi_{\mathbf{R}}(\hat{\mathbf{v}}_{k})$ for
$k=1,2,\dots,k_{0}$. (If $k_{0}=0$, we have
$\alpha(\beta^{\prime})=\mathbb{L}({\mathbf{S}})^{\infty}$ and
$\alpha(\hat{\beta})=\mathbb{L}(\mathbf{R}^{\prime})^{\infty}$.)

The words $\hat{\mathbf{v}}_{1},\hat{\mathbf{v}}_{2},\dots$ determine a
(finite or infinite) collection
$\widehat{\mathcal{I}}=\\{\hat{I}_{1},\hat{I}_{2},\dots\\}$ of intervals in
the same way that $\mathbf{v}_{1},\mathbf{v}_{2},\dots$ determine
$I_{1},I_{2},\dots$. In fact, if
$\hat{I}_{k}=[\underline{\hat{t}}_{k},\overline{\hat{t}}_{k})$, then

(12.1)
$b(\underline{t}_{k},\beta)=\Phi_{\mathbf{R}}\big{(}b(\underline{\hat{t}}_{k},\hat{\beta})\big{)},\qquad
b(\overline{t}_{k},\beta)=\Phi_{\mathbf{R}}\big{(}b(\overline{\hat{t}}_{k},\hat{\beta})\big{)}.$

Since $\mathbf{w}^{\infty}\succ\mathbf{R}^{-}\mathbb{L}(\mathbf{R})^{\infty}$,
we have $\mathbf{w}=\Phi_{\mathbf{R}}(\hat{\mathbf{w}})$ for some
$\hat{\mathbf{w}}\in L^{*}(\hat{\beta})$, as in the proof of Proposition 7.8,
and so there is a point $\hat{t}_{R}$ such that
$b(\hat{t}_{R},\hat{\beta})=\hat{\mathbf{w}}^{\infty}$. Therefore, since
$t_{R}^{\prime}\not\in\bigcup_{k=1}^{k_{0}}I_{k}^{\prime}$ and
$b(t_{R}^{\prime},\beta^{\prime})=\mathbf{w}^{\infty}=\Phi_{\mathbf{R}}(\hat{\mathbf{w}}^{\infty})=\Phi_{\mathbf{R}}(b(\hat{t}_{R},\hat{\beta}))$,
it follows using (12.1) that
$\hat{t}_{R}\not\in\bigcup_{k=1}^{k_{0}}\hat{I}_{k}$. Furthermore, since