Proof. Note that $01=U_{1}(0)=U_{0}(1)\in F_{1}^{*}\subset \Omega _{F}^{*}$ . Furthermore, for $p\in \mathbb {N}$ and $0 t_{1}\ldots t_{N}1\in \Omega _{F}^{*}$ , we have

$$ \begin{align*} 01^{p}=U_{1}(01^{p-1})&=U_{1}^{p-1}(U_{0}(1)),\\[3pt] 0^{p}1=U_{0}(0^{p-1}1)&=U_{0}^{p-1}(U_{1}(0)),\\[3pt] 01^{p} 01^{p+t_{1}}\ldots 01^{p+t_{N}}01^{p+1}&=U_{1}^{p}(U_{0}(0t_{1}\ldots t_{N}1)),\\[3pt] 0^{p+1}10^{p+t_{1}}1\ldots 0^{p+t_{N}}10^{p} 1&=U_{0}^{p}(U_{1}(0\,\overline{t_{1}\ldots t_{N}}\,1)). \end{align*} $$

By Lemma 2.3(iii), if $0 t_{1}\ldots t_{N}1\in \Omega _{F}^{*}$ , then $0\,\overline {t_{1}\ldots t_{N}}\,1\in \Omega _{F}^{*}$ as well. Hence by Lemma 2.2, all the above words lie in $\Omega _{F}^{*}$ .

To prove the converse, it suffices to show that each word in $\Omega _{F}^{*}$ is of the form $U_{0}^{p}(U_{1}( {\mathbf {t}}))$ or $U_{1}^{p}(U_{0}( {\mathbf {t}}))$ for some $p\geq 0$ and Farey word $ {\mathbf {t}}\in \Omega _{F}$ . This is clearly true for $01=U_{0}^{0}(U_{1}(0))$ , where $U_{0}^{0}$ denotes the identity map. Let $n\geq 1$ and suppose the statement is true for all Farey words in $F_{n}^{*}$ . Take $ {\mathbf {s}}\in F_{n+1}^{*}$ with $ {\mathbf {s}}\neq 01$ . By Lemmas 2.1 and 2.2, $ {\mathbf {s}}=U_{0}( {\mathbf {t}})$ or $ {\mathbf {s}}=U_{1}( {\mathbf {t}})$ for some Farey word $ {\mathbf {t}}\in F_{n}^{*}$ . We assume the former, as the argument for the second case is similar. By the induction hypothesis, either $ {\mathbf {t}}=U_{0}^{p}(U_{1}(\mathbf {u}))$ for some $\mathbf {u}\in \Omega _{F}$ and $p\geq 0$ , in which case $ {\mathbf {s}}=U_{0}^{p+1}(U_{1}(\mathbf {u}))$ ; or $ {\mathbf {t}}=U_{1}^{p}(U_{0}(\mathbf {u}))$ for some $\mathbf {u}\in \Omega _{F}$ and $p\geq 1$ , in which case $ {\mathbf {s}}=U_{0}(U_{1}(\mathbf {v}))$ , where $\mathbf {v}=U_{1}^{p-1}(U_{0}(\mathbf {u}))\in \Omega _{F}$ . In both cases, $ {\mathbf {s}}$ is of the required form.

Proof. First we prove equation (2.4). Since $ {\mathbf {s}}$ is Lyndon, it is not periodic. Hence $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ is not periodic, because any cyclic permutation of a periodic word is periodic. Since $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ , we have

$$ \begin{align*} a_{i+1}\ldots a_{m}\preccurlyeq a_{1}\ldots a_{m-i}\quad\text{for all } 1\le i<m. \end{align*} $$

Suppose equality holds for some i. Then

$$ \begin{align*} a_{i+1}\ldots a_{m} a_{1}\ldots a_{i}=a_{1}\ldots a_{m-i} a_{1}\ldots a_{i}\succcurlyeq a_{1}\ldots a_{m-i} a_{m-i+1}\ldots a_{m}= \mathbf{a}, \end{align*} $$

so $a_{i+1}\ldots a_{m} a_{1}\ldots a_{i}=\mathbb {L}( {\mathbf {s}})= \mathbf {a}$ by definition of $\mathbb {L}( {\mathbf {s}})$ . By Lemma 2.6, this cannot happen, since $ \mathbf {a}$ is not periodic.

Next we prove equation (2.5). Since $ {\mathbf {s}}=s_{1}\ldots s_{m}$ is a Lyndon word, any word of length $k\in \{1, \ldots , m-1\}$ occurring in $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ is lexicographically larger than or equal to $s_{1}\ldots s_{k}$ . By equation (2.4), it follows that

(2.6) $$ \begin{align} a_{k+1}\ldots a_{m}^{+} s_{1}\ldots s_{k}\preccurlyeq a_{1}\ldots a_{m-k}a_{m-k+1}\ldots a_{m}\prec a_{1}\ldots a_{m}^{+} \end{align} $$

for all $0<k<m$ . Hence, by equations (2.6) and (2.4), we conclude that $\sigma ^{n}( \mathbf {a}^{+}{ {\mathbf {s}}^{-}} \mathbf {a}^{\infty })\prec \mathbf {a}^{+}{ {\mathbf {s}}^{-}} \mathbf {a}^{\infty }$ for all $n\ge 1$ . This completes the proof.

Proof. Suppose $\delta (\beta )=(a_{1}\ldots a_{m})^{\infty }$ with minimal period block $ \mathbf {a}=a_{1}\ldots a_{m}$ . Then $m\ge 2$ since $\beta <2$ . Take $ {\mathbf {s}}:=\mathbb {S}( \mathbf {a})$ . Then $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ , and

(2.7) $$ \begin{align} s_{i+1}\ldots s_{m}\succcurlyeq s_{1}\ldots s_{m-i} \quad\text{for all}\ 1\leq i<m. \end{align} $$

If equality holds for some i, then we deduce just as in the proof of Lemma 2.7 that $ {\mathbf {s}}$ is periodic. However, then $ \mathbf {a}$ is also periodic, contradicting that m is the minimal period of $\delta (\beta )$ . Hence, strict inequality holds in equation (2.7), and $ {\mathbf {s}}$ is Lyndon. The converse is trivial.

Proof. Let $(d_{i})$ and $(d_{i}^{\prime })$ be two sequences in $X_{E}$ , and let $(e_{i}), (e_{i}^{\prime })$ be their corresponding edge paths; thus, $(d_{i})=\mathcal L^{E}((e_{i}))$ and $(d_{i}^{\prime })=\mathcal L^{E}((e_{i}^{\prime }))$ . Suppose $(d_{i})\prec (d_{i}^{\prime })$ . Then there exists $k\in \mathbb {N}$ such that $d_{1}\ldots d_{k-1}=d_{1}^{\prime }\ldots d_{k-1}^{\prime }$ and $d_{k}<d_{k}^{\prime }$ . If $k=1$ , then $d_{1}=0$ and $d_{1}^{\prime }=1$ . So, $\mathcal L^{V}(t(e_{1}))={ {\mathbf {s}}^{-}}$ and $\mathcal L^{V}(t(e_{1}^{\prime }))= \mathbf {a}^{+}$ . By Definition 3.1, it follows that $\Phi _ {\mathbf {s}}((d_{i}))\prec \Phi _ {\mathbf {s}}((d_{i}^{\prime }))$ .

If $k>1$ , then $e_{1}\ldots e_{k-1}=e_{1}^{\prime }\ldots e_{k-1}^{\prime }$ , which implies that the initial vertices of $e_{k}$ and $e_{k}^{\prime }$ coincide. Since $d_{k}<d_{k}^{\prime }$ , by the definition of $\mathcal L^{V}$ , it follows that (see Figure 2)

$$ \begin{align*} \mathcal L^{V}(t(e_{k}))\prec\mathcal L^{V}(t(e_{k}^{\prime})). \end{align*} $$

By Definition 3.1, we also have $\Phi _ {\mathbf {s}}((d_{i}))\prec \Phi _ {\mathbf {s}}((d_{i}^{\prime }))$ . This completes the proof.

Proof. Since $ {\mathbf {d}}=d_{1}\ldots d_{k}\in B_{*}(X_{E})$ , there exists a unique finite edge path $e_{1}\ldots e_{k}$ such that $\mathcal L^{E}(e_{1}\ldots e_{k})= {\mathbf {d}}$ . If $d_{k}=1$ , then $ {\mathbf {d}}^{-}=d_{1}\ldots d_{k-1}0$ can be represented by a unique finite edge path $e_{1}^{\prime }\ldots e_{k}^{\prime }$ with $e_{1}^{\prime }\ldots e_{k-1}^{\prime }=e_{1}\ldots e_{k-1}$ . By the definition of $\mathcal L^{V}$ , it follows that $\mathcal L^{V}(t(e_{k}^{\prime }))=\mathcal L^{V}(t(e_{k}))^{-}$ . Therefore, by Definition 3.1, it follows that

$$ \begin{align*} \Phi_ {\mathbf{s}}( {\mathbf{d}}^{-})=\Phi_ {\mathbf{s}}(\mathcal L^{E}(e_{1}^{\prime}\ldots e_{k}^{\prime})) &= \Phi_ {\mathbf{s}}(\mathcal L^{E}(e_{1}\ldots e_{k-1}e_{k}^{\prime}))\\[3pt] &=\mathcal L^{V}(t(e_{1})\ldots t(e_{k-1})t(e_{k}^{\prime})) \\[3pt] &=\mathcal L^{V}(t(e_{1})\ldots t(e_{k}))^{-}=\Phi_ {\mathbf{s}}( {\mathbf{d}})^{-}. \end{align*} $$

This proves the first equality of the lemma. The second equality follows analogously.

Proof. Let $ {\mathbf {s}}=s_{1}\ldots s_{m}\in \Omega _{L}^{*}$ and $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ . Then there exists $j\in \{1,\ldots ,m-1\}$ such that

(3.3) $$ \begin{align} \mathbf{a}=s_{j+1}\ldots s_{m} s_{1}\ldots s_{j}. \end{align} $$

Let $ {\mathbf {r}}=r_{1}\ldots r_{\ell }\in \Omega _{L}^{*}$ . Then we can write $ {\mathbf {s}}\bullet {\mathbf {r}}=\Phi _ {\mathbf {s}}( {\mathbf {r}})=b_{1}\ldots b_{m\ell }$ . Furthermore, there exists a finite edge path $e_{1}\ldots e_{\ell }$ representing $ {\mathbf {r}}$ such that

$$ \begin{align*}\Phi_ {\mathbf{s}}( {\mathbf{r}})=\mathcal L^{V}(t(e_{1}) \ldots t(e_{\ell}))=: {\mathbf{b}}_{1}\ldots {\mathbf{b}}_{\ell},\end{align*} $$

where each block $ {\mathbf {b}}_{i}\in \{{ {\mathbf {s}}^{-}}, {\mathbf {s}}, \mathbf {a}, \mathbf {a}^{+}\}$ . Note that $ {\mathbf {b}}_{1}={ {\mathbf {s}}^{-}}$ since the block $ {\mathbf {r}}$ begins with $r_{1}=0$ . By Definition 1.1, it suffices to prove

(3.4) $$ \begin{align} b_{i+1}\ldots b_{m\ell}\succ b_{1}\ldots b_{m\ell-i}\quad\textrm{for any }0<i<m\ell. \end{align} $$

We split the proof of equation (3.4) into two cases.

Case I. $i=km$ for some $k\in \{1, 2,\ldots , \ell -1\}$ . Then $b_{i+1}\ldots b_{m\ell }= {\mathbf {b}}_{k+1}\ldots {\mathbf {b}}_{\ell }$ . Since $ {\mathbf {r}}$ is a Lyndon word, we have $r_{k+1}\ldots r_{\ell }\succ r_{1}\ldots r_{\ell -k}$ . So, equation (3.4) follows directly by Lemma 3.3.

Case II. $i=km+p$ for some $k\in \{0,1,\ldots , \ell -1\}$ and $p\in \{1,\ldots , m-1\}$ . Then $b_{i+1}\ldots b_{m\ell }=b_{i+1}\ldots b_{i+m-p} {\mathbf {b}}_{k+2}\ldots {\mathbf {b}}_{\ell }$ . In the following, we prove equation (3.4) by considering the four possible choices of $ {\mathbf {b}}_{k+1}\in \{{ {\mathbf {s}}^{-}}, {\mathbf {s}}, \mathbf {a}, \mathbf {a}^{+}\}$ . If $ {\mathbf {b}}_{k+1}= {\mathbf {s}}$ , then by using that $ {\mathbf {s}}\in \Omega _{L}^{*}$ , we conclude that

$$ \begin{align*}b_{i+1}\ldots b_{i+m-p}=s_{p+1}\ldots s_{m}\succ s_{1}\ldots s_{m-p}=b_{1}\ldots b_{m-p},\end{align*} $$

proving equation (3.4). Similarly, if $ {\mathbf {b}}_{k+1}= \mathbf {a}^{+}$ , then by equation (3.3), one can also prove that $b_{i+1}\ldots b_{i+m-p} \succ b_{1}\ldots b_{m-p}$ . Now we assume $ {\mathbf {b}}_{k+1}={ {\mathbf {s}}^{-}}$ . Then by using $ {\mathbf {s}}\in \Omega _{L}^{*}$ , it follows that

(3.5) $$ \begin{align} b_{i+1}\ldots b_{i+m-p}=s_{p+1}\ldots s_{m}^{-}\succcurlyeq s_{1}\ldots s_{m-p}=b_{1}\ldots b_{m-p}. \end{align} $$

Observe that the word ${ {\mathbf {s}}^{-}}$ can only be followed by $ \mathbf {a}$ or $ \mathbf {a}^{+}$ in G (see Figure 2). So $ {\mathbf {b}}_{k+2}\in \{ \mathbf {a}, \mathbf {a}^{+}\}$ . Since $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})$ , we obtain that

(3.6) $$ \begin{align} b_{i+m-p+1}\ldots b_{i+m}=a_{1}\ldots a_{p}\succcurlyeq s_{m-p+1}\ldots s_{m}\succ s_{m-p+1}\ldots s_{m}^{-}=b_{m-p+1}\ldots b_{m}. \end{align} $$

Thus, by equations (3.5) and (3.6), we conclude that $b_{i+1}\ldots b_{i+m}\succ b_{1}\ldots b_{m}$ , proving equation (3.4). Finally, suppose $ {\mathbf {b}}_{k+1}= \mathbf {a}$ . Note that the word $ \mathbf {a}$ can only be followed by $ \mathbf {a}$ or $ \mathbf {a}^{+}$ in G. Then by equation (3.3) and using $ {\mathbf {s}}\in \Omega _{L}^{*}$ , we have

$$ \begin{align*} b_{i+1}\ldots b_{i+m}\succcurlyeq s_{1}\ldots s_{m}\succ b_{1}\ldots b_{m}. \end{align*} $$

This completes the proof.

Proof. To prove (i), it suffices to show that if $ {\mathbf {b}}_{1}, {\mathbf {b}}_{2}$ is a connectible sequence, then $\Phi _ {\mathbf {s}}( {\mathbf {b}}_{1} {\mathbf {b}}_{2})=\Phi _ {\mathbf {s}}( {\mathbf {b}}_{1})\Phi _ {\mathbf {s}}( {\mathbf {b}}_{2})$ ; the statement then extends to arbitrary connectible sequences by induction.

Without loss of generality, by the symmetry of the edge-labels in Figure 2, we may assume that $ {\mathbf {b}}_{1}$ ends in the digit 0 and $ {\mathbf {b}}_{2}$ begins with the digit 1. However, note that in the directed graph in Figure 2, if we travel along an edge labeled 0 followed by an edge labeled 1, we always end up at the vertex labeled $ \mathbf {a}^{+}$ , which is also the first vertex visited after traveling along an edge labeled 1 from the ‘Start-1’ vertex. Thus, $\Phi _ {\mathbf {s}}( {\mathbf {b}}_{1} {\mathbf {b}}_{2})=\Phi _ {\mathbf {s}}( {\mathbf {b}}_{1})\Phi _ {\mathbf {s}}( {\mathbf {b}}_{2})$ .

Statement (ii) follows from (i) since $ {\mathbf {r}}$ begins with digit 0 and ends with digit 1, so $ {\mathbf {r}}$ is connectible to itself; and similarly, $\mathbb {L}( {\mathbf {r}})$ begins with digit 1 and ends with digit 0, so $\mathbb {L}( {\mathbf {r}})$ is connectible to itself.

Proof. The proof is similar to that of Lemma 3.5. Let $ {\mathbf {r}}=r_{1}\ldots r_{\ell }\in \Omega _{L}^{*}$ . First we show that $ {\mathbf {s}}\bullet \mathbb {L}( {\mathbf {r}})$ is a cyclic permutation of $ {\mathbf {s}}\bullet {\mathbf {r}}$ . Note that $\mathbb {L}( {\mathbf {r}})=r_{j+1}\ldots r_{\ell } r_{1}\ldots r_{j}$ for some $1<j<\ell $ . Then $r_{j}=0$ and $r_{j+1}=1$ , so Lemma 3.6(i) implies that

(3.7) $$ \begin{align} {\mathbf{s}}\bullet {\mathbf{r}}=\Phi_ {\mathbf{s}}(r_{1}\ldots r_{\ell})=\Phi_ {\mathbf{s}}(r_{1}\ldots r_{j})\Phi_ {\mathbf{s}}(r_{j+1}\ldots r_{\ell}). \end{align} $$

However, since $ {\mathbf {r}}\in \Omega _{L}^{*}$ , we have $r_{\ell }=1$ and $r_{1}=0$ , so by Lemma 3.6(i), we obtain that

$$ \begin{align*} {\mathbf{s}}\bullet\mathbb{L}( {\mathbf{r}})=\Phi_ {\mathbf{s}}(r_{j+1}\ldots r_{\ell} r_{1}\ldots r_{j})=\Phi_ {\mathbf{s}}(r_{j+1}\ldots r_{\ell})\Phi_ {\mathbf{s}}(r_{1}\ldots r_{j}). \end{align*} $$

This, together with equation (3.7), proves that $ {\mathbf {s}}\bullet \mathbb {L}( {\mathbf {r}})$ is indeed a cyclic permutation of $ {\mathbf {s}}\bullet {\mathbf {r}}$ . It remains to prove that $ {\mathbf {s}}\bullet \mathbb {L}( {\mathbf {r}})$ is the lexicographically largest cyclic permutation of itself.

Write $ {\mathbf {s}}=s_{1}\ldots s_{m}\in \Omega _{L}^{*}$ with $ \mathbf {a}=\mathbb {L}( {\mathbf {s}})=a_{1}\ldots a_{m}$ , and write $\mathbb {L}( {\mathbf {r}})=c_{1}\ldots c_{\ell }$ . Then by Lemma 2.7, it follows that

(3.8) $$ \begin{align} a_{i+1}\ldots a_{m}&\prec a_{1}\ldots a_{m-i}\quad\text{for all } 0<i<m;\nonumber\\[5pt] c_{i+1}\ldots c_{\ell}&\prec c_{1}\ldots c_{\ell-i}\quad \text{for all } 0<i<\ell. \end{align} $$

Write $ {\mathbf {s}}\bullet \mathbb {L}( {\mathbf {r}})= {\mathbf {b}}_{1}\ldots {\mathbf {b}}_{\ell }=b_{1}\ldots b_{m\ell }$ , where each $ {\mathbf {b}}_{i}\in \{{ {\mathbf {s}}^{-}}, {\mathbf {s}}, \mathbf {a}, \mathbf {a}^{+}\}$ . Then it suffices to prove that

(3.9) $$ \begin{align} b_{i+1}\ldots b_{m\ell}\prec b_{1}\ldots b_{m\ell-i}\quad\text{for all } 0<i<m\ell. \end{align} $$

Since $\mathbb {L}( {\mathbf {r}})$ has a prefix $c_{1}=1$ , we see that $b_{1}\ldots b_{m}= {\mathbf {b}}_{1}= \mathbf {a}^{+}$ . So, by using equation (3.8) and the same argument as in the proof of Lemma 3.5, we can prove equation (3.9).

Proof. (i) Suppose $d_{1}=1$ and $\sigma ^{n}((c_{i}))\prec (d_{i})$ for all $n\ge 0$ . Then $\Phi _{ {\mathbf {s}}}((d_{i}))$ begins with $\mathbb {L}( {\mathbf {s}})^{+}$ . If $n\equiv 0\pmod {| {\mathbf {s}}|}$ , then by Lemma 3.3, it follows that $\sigma ^{n}(\Phi _ {\mathbf {s}}((c_{i})))\prec \Phi _ {\mathbf {s}}((d_{i}))$ . If $n\ne 0\pmod {| {\mathbf {s}}|}$ , then by using $\Phi _ {\mathbf {s}}(d_{1})=\mathbb {L}( {\mathbf {s}})^{+}$ and the same argument as in the proof of Lemma 3.7, one can verify that $\sigma ^{n}(\Phi _ {\mathbf {s}}((c_{i})))\prec \Phi _ {\mathbf {s}}((d_{i}))$ . The proof of (ii) is similar.

Proof. Let $ {\mathbf {r}}=r_{1}\ldots r_{m}, {\mathbf {s}}=s_{1}\ldots s_{n}$ , and $ {\mathbf {t}}=t_{1}\ldots t_{\ell }$ . Then we can write $( {\mathbf {r}}\bullet {\mathbf {s}})\bullet {\mathbf {t}}$ as

(3.10) $$ \begin{align} ( {\mathbf{r}}\bullet {\mathbf{s}})\bullet {\mathbf{t}}=\mathbf B_{1}\mathbf B_{2}\ldots \mathbf B_{\ell}, \end{align} $$

where each $\mathbf B_{i}\in \{( {\mathbf {r}}\bullet {\mathbf {s}})^{-}, {\mathbf {r}}\bullet {\mathbf {s}}, \mathbb {L}( {\mathbf {r}}\bullet {\mathbf {s}}), \mathbb {L}( {\mathbf {r}}\bullet {\mathbf {s}})^{+}\}$ . Since $t_{1}=0$ , we have ${\mathbf B_{1}=( {\mathbf {r}}\bullet {\mathbf {s}})^{-}}$ . Furthermore, by the definition of $\Phi _{ {\mathbf {r}}\bullet {\mathbf {s}}}$ it follows that for $1<i\le \ell $ ,

(3.11) $$ \begin{align} \mathbf B_{i}=\left\{ \begin{array} {@{}ll} \mathbb{L}( {\mathbf{r}}\bullet {\mathbf{s}})&\textrm{if } t_{i-1}t_{i}=00,\\[3pt] \mathbb{L}( {\mathbf{r}}\bullet {\mathbf{s}})^{+}&\textrm{if } t_{i-1}t_{i}=01,\\[3pt] ( {\mathbf{r}}\bullet {\mathbf{s}})^{-}&\textrm{if } t_{i-1}t_{i}=10,\\[3pt] {\mathbf{r}}\bullet {\mathbf{s}}&\textrm{if } t_{i-1}t_{i}=11. \end{array} \right. \end{align} $$

Similarly, we can write

$$ \begin{align*} {\mathbf{s}}\bullet {\mathbf{t}}= {\mathbf{b}}_{1} {\mathbf{b}}_{2}\ldots {\mathbf{b}}_{\ell}, \end{align*} $$

where each $ {\mathbf {b}}_{i}\in \{{ {\mathbf {s}}^{-}}, {\mathbf {s}}, \mathbb {L}( {\mathbf {s}}), \mathbb {L}( {\mathbf {s}})^{+}\}$ , and it follows from the definition of $\Phi _ {\mathbf {s}}$ that

(3.12) $$ \begin{align} {\mathbf{b}}_{i}=\left\{ \begin{array} {@{}ll} \mathbb{L}( {\mathbf{s}})&\textrm{if } t_{i-1}t_{i}=00,\\[3pt] \mathbb{L}( {\mathbf{s}})^{+}&\textrm{if } t_{i-1}t_{i}=01,\\[3pt] { {\mathbf{s}}^{-}}&\textrm{if } t_{i-1}t_{i}=10,\\[3pt] {\mathbf{s}}&\textrm{if } t_{i-1}t_{i}=11, \end{array} \right. \end{align} $$

for $1<i\le \ell $ . Comparing equations (3.11) and (3.12) and using Lemmas 3.4 and 3.7, it follows that

$$ \begin{align*}\Phi_ {\mathbf{r}}( {\mathbf{b}}_{i})=\mathbf{B}_{i}\quad\text{for all } i\ge 1.\end{align*} $$

Moreover, the sequence $ {\mathbf {b}}_{1}, {\mathbf {b}}_{2},\ldots , {\mathbf {b}}_{\ell }$ is connectible because $ {\mathbf {b}}_{1} {\mathbf {b}}_{2}\ldots {\mathbf {b}}_{\ell }=\Phi _ {\mathbf {s}}( {\mathbf {t}})$ arises from a walk along the directed graph in Figure 2. Hence, Lemma 3.6 and equation (3.10) yield

$$ \begin{align*} {\mathbf{r}}\bullet( {\mathbf{s}}\bullet {\mathbf{t}})=\Phi_ {\mathbf{r}}( {\mathbf{b}}_{1} {\mathbf{b}}_{2}\ldots {\mathbf{b}}_{\ell})=\Phi_ {\mathbf{r}}( {\mathbf{b}}_{1})\Phi_ {\mathbf{r}}( {\mathbf{b}}_{2})\ldots\Phi_ {\mathbf{r}}( {\mathbf{b}}_{\ell})=\mathbf{B}_{1}\mathbf{B}_{2}\ldots\mathbf{B}_{\ell}=( {\mathbf{r}}\bullet {\mathbf{s}})\bullet {\mathbf{t}}, \end{align*} $$

as desired.

Proof of Proposition 1.3

The proposition follows by Lemmas 3.5 and 3.9 and Example 1.2, which shows that $\bullet $ is not commutative.

Proof. Let $I^ {\mathbf {S}}=[\beta _{\ell }^ {\mathbf {S}}, \beta _{*}^ {\mathbf {S}}]$ and $J^{ {\mathbf {S}}}=[\beta _{\ell }^ {\mathbf {S}}, \beta _{r}^ {\mathbf {S}}]$ . Then by equation (5.1), it follows that

$$ \begin{align*} \delta(\beta_{*}^ {\mathbf{S}})=\mathbb{L}( {\mathbf{S}})^{+} {\mathbf{S}}^{-}\mathbb{L}( {\mathbf{S}})^{\infty}\prec \mathbb{L}( {\mathbf{S}})^{+} {\mathbf{S}}^{\infty}=\delta(\beta_{r}^ {\mathbf{S}}), \end{align*} $$

which implies $\beta _{*}^ {\mathbf {S}}<\beta _{r}^ {\mathbf {S}}$ by Lemma 2.5. So $I^ {\mathbf {S}} \subset J^ {\mathbf {S}}$ .

For (i), let $ {\mathbf {r}}\in \Omega _{F}^{*}$ . Then $ {\mathbf {r}}$ begins with digit $0$ and ends with digit $1$ . By Lemmas 3.6(ii) and 3.7, this implies that

$$ \begin{align*} \delta(\beta_{\ell}^{ {\mathbf{S}}\bullet {\mathbf{r}}})&= \mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{\infty}=( {\mathbf{S}}\bullet\mathbb{L}( {\mathbf{r}}))^{\infty}=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{\infty})\\[3pt] &\succ \Phi_ {\mathbf{S}}(10^{\infty})=\mathbb{L}( {\mathbf{S}})^{+} {\mathbf{S}}^{-}\mathbb{L}( {\mathbf{S}})^{\infty}=\delta(\beta_{*}^ {\mathbf{S}}). \end{align*} $$

So, $\beta _{\ell }^{ {\mathbf {S}}\bullet {\mathbf {r}}}>{\beta _{*}^ {\mathbf {S}}}$ . Furthermore, by Lemmas 3.4, 3.6(ii), and 3.7, it follows that

$$ \begin{align*} \delta(\beta_{r}^{ {\mathbf{S}}\bullet {\mathbf{r}}})&=\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{+}( {\mathbf{S}}\bullet {\mathbf{r}})^{\infty}\\[3pt] &=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{+})\Phi_ {\mathbf{S}}( {\mathbf{r}}^{\infty})=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{+} {\mathbf{r}}^{\infty})\\[3pt] &\prec \Phi_ {\mathbf{S}}(1^{\infty})=\mathbb{L}( {\mathbf{S}})^{+} {\mathbf{S}}^{\infty}=\delta(\beta_{r}^ {\mathbf{S}}). \end{align*} $$

This proves $\beta _{r}^{ {\mathbf {S}}\bullet {\mathbf {r}}}<\beta _{r}^ {\mathbf {S}}$ . Hence, $J^{ {\mathbf {S}}\bullet {\mathbf {r}}}=[\beta _{\ell }^{ {\mathbf {S}}\bullet {\mathbf {r}}}, \beta _{r}^{ {\mathbf {S}}\bullet {\mathbf {r}}}]\subset (\beta _{*}^ {\mathbf {S}}, \beta _{r}^ {\mathbf {S}}]=J^ {\mathbf {S}}\setminus I^ {\mathbf {S}}$ .

Next we prove (ii). Let $ {\mathbf {r}}, {\mathbf {r}}^{\prime }$ be two different Farey words in $\Omega _{F}^{*}$ . By Lemma 2.9(i), it follows that $J^{ {\mathbf {r}}}\cap J^{ {\mathbf {r}}^{\prime }}=\emptyset $ . Write $J^{ {\mathbf {r}}}=[\beta _{\ell }^ {\mathbf {r}}, \beta _{r}^ {\mathbf {r}}]$ and $J^{ {\mathbf {r}}^{\prime }}=[\beta _{\ell }^{ {\mathbf {r}}^{\prime }}, \beta _{r}^{ {\mathbf {r}}^{\prime }}]$ . Since $J^ {\mathbf {r}}$ and $J^{ {\mathbf {r}}^{\prime }}$ are disjoint, we may assume $\beta _{r}^ {\mathbf {r}}<\beta _{\ell }^{ {\mathbf {r}}^{\prime }}$ . By equation (5.1) and Lemma 2.5, it follows that

(5.2) $$ \begin{align} \mathbb{L}( {\mathbf{r}})^{+} {\mathbf{r}}^{\infty}=\delta(\beta_{r}^ {\mathbf{r}})\prec \delta(\beta_{\ell}^{ {\mathbf{r}}^{\prime}})=\mathbb{L}( {\mathbf{r}}^{\prime})^{\infty}. \end{align} $$

Then by equations (5.1), (5.2), and Lemma 3.3, we obtain that

$$ \begin{align*} \delta(\beta_{r}^{ {\mathbf{S}}\bullet {\mathbf{r}}})&=\mathbb{L}(S\bullet {\mathbf{r}})^{+}(S\bullet {\mathbf{r}})^{\infty}=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{+} {\mathbf{r}}^{\infty})\\[3pt] &\prec\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}}^{\prime})^{\infty})=\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}}^{\prime})^{\infty}=\delta(\beta_{\ell}^{ {\mathbf{S}}\bullet {\mathbf{r}}^{\prime}}). \end{align*} $$

It follows that $\beta _{r}^{ {\mathbf {S}}\bullet {\mathbf {r}}}<\beta _{\ell }^{ {\mathbf {S}}\bullet {\mathbf {r}}^{\prime }}$ , and hence $J^{ {\mathbf {S}}\bullet {\mathbf {r}}}\cap J^{ {\mathbf {S}}\bullet {\mathbf {r}}^{\prime }}=\emptyset $ .

Proof. Let $ {\mathbf {S}}\in \Lambda $ with $ {\mathbf {A}}=\mathbb {L}( {\mathbf {S}})$ . First we show that the map $\Psi _ {\mathbf {S}}: (1,2]\to J^ {\mathbf {S}}\setminus I^ {\mathbf {S}}=(\beta _{*}^ {\mathbf {S}}, \beta _{r}^ {\mathbf {S}}]$ is well defined. Note that

(5.4) $$ \begin{align} \delta(\beta_{*}^ {\mathbf{S}})= {\mathbf{A}}^{+} {\mathbf{S}}^{-} {\mathbf{A}}^{\infty}=\Phi_ {\mathbf{S}}(10^{\infty})\quad\textrm{and}\quad \delta(\beta_{r}^ {\mathbf{S}})= {\mathbf{A}}^{+} {\mathbf{S}}^{\infty}=\Phi_ {\mathbf{S}}(1^{\infty}). \end{align} $$

Take $\beta \in (1,2]$ . Then $10^{\infty }\prec \delta (\beta )\preccurlyeq 1^{\infty }$ . By Lemma 3.3 and equation (5.4), it follows that

$$ \begin{align*} \delta(\beta_{*}^ {\mathbf{S}})\prec\Phi_ {\mathbf{S}}(\delta(\beta))\preccurlyeq \delta(\beta_{r}^ {\mathbf{S}}). \end{align*} $$

Thus, by Lemma 2.5, it suffices to prove that

(5.5) $$ \begin{align} \sigma^{n}(\Phi_ {\mathbf{S}}(\delta(\beta)))\preccurlyeq\Phi_ {\mathbf{S}}(\delta(\beta))\quad\text{for all } n\ge 0. \end{align} $$

Note by Lemma 2.5 that $\sigma ^{n}(\delta (\beta ))\preccurlyeq \delta (\beta )$ for all $n\ge 0$ , and $\delta (\beta )$ begins with digit $1$ . Thus, equation (5.5) follows by Lemma 3.8(i), and we conclude that the map $\Psi _ {\mathbf {S}}$ is well defined.

The monotonicity of $\Psi _ {\mathbf {S}} =\delta ^{-1}\circ \Phi _ {\mathbf {S}}\circ \delta $ follows since both maps $\delta $ and $\Phi _ {\mathbf {S}}$ are strictly increasing by Lemmas 2.5 and 3.3, respectively. This completes the proof.

Proof. We first prove that

(5.6) $$ \begin{align} \Psi_ {\mathbf{S}}(\beta_{\ell}^ {\mathbf{r}})=\beta_{\ell}^{ {\mathbf{S}}\bullet {\mathbf{r}}}\quad\textrm{and}\quad \Psi_ {\mathbf{S}}(\beta_{r}^ {\mathbf{r}})=\beta_{r}^{ {\mathbf{S}}\bullet {\mathbf{r}}} \quad\text{for all } {\mathbf{r}}\in{\Omega_{L}^{*}}. \end{align} $$

Observe that $\delta (\beta _{\ell }^ {\mathbf {r}})=\mathbb {L}( {\mathbf {r}})^{\infty }$ . Then by Lemmas 3.6(ii) and 3.7, it follows that

$$ \begin{align*} \Phi_ {\mathbf{S}}(\delta(\beta_{\ell}^ {\mathbf{r}}))=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{\infty})=( {\mathbf{S}}\bullet\mathbb{L}( {\mathbf{r}}))^{\infty}=\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{\infty}=\delta(\beta_{\ell}^{ {\mathbf{S}}\bullet {\mathbf{r}}}), \end{align*} $$

so $\Psi _ {\mathbf {S}}(\beta _{\ell }^ {\mathbf {r}})=\beta _{\ell }^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ . Similarly, since $\delta (\beta _{r}^ {\mathbf {r}})=\mathbb {L}( {\mathbf {r}})^{+} {\mathbf {r}}^{\infty }$ , Lemmas 3.4, 3.6(ii), and 3.7 imply that

$$ \begin{align*} \Phi_ {\mathbf{S}}(\delta(\beta_{r}^ {\mathbf{r}}))&=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{+} {\mathbf{r}}^{\infty})=\Phi_ {\mathbf{S}}(\mathbb{L}( {\mathbf{r}})^{+})\Phi_ {\mathbf{S}}( {\mathbf{r}}^{\infty})\\[3pt] &=( {\mathbf{S}}\bullet\mathbb{L}( {\mathbf{r}}))^{+}\Phi_ {\mathbf{S}}( {\mathbf{r}})^{\infty}=\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{+}( {\mathbf{S}}\bullet {\mathbf{r}})^{\infty}=\delta(\beta_{r}^{ {\mathbf{S}}\bullet {\mathbf{r}}}). \end{align*} $$

We conclude that $\Psi _ {\mathbf {S}}(\beta _{r}^{ {\mathbf {r}}})=\beta _{r}^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ . This proves equation (5.6).

Note by Lemma 2.5(ii) that the map $\beta \mapsto \delta (\beta )$ is left continuous in $(1,2]$ , and is right continuous at a point $\beta _{0}$ if and only if $\delta (\beta _{0})$ is not periodic. Hence by Lemma 2.8, it follows that the map $\beta \mapsto \delta (\beta )$ is continuous at $\beta _{0}$ if and only if $\delta (\beta _{0})$ is not of the form $\mathbb {L}( {\mathbf {r}})^{\infty }$ for a Lyndon word $ {\mathbf {r}}$ . Since $\Phi _ {\mathbf {S}}$ is clearly continuous with respect to the order topology and the map $\delta ^{-1}$ is continuous, it follows that $\Psi _ {\mathbf {S}}$ is continuous at $\beta _{0}$ if and only if $\delta (\beta _{0})$ is not of the form $\mathbb {L}( {\mathbf {r}})^{\infty }$ for a Lyndon word $ {\mathbf {r}}$ . Moreover, $\Psi _ {\mathbf {S}}$ is left continuous everywhere.

Note that if $\delta (\beta _{0})=\mathbb {L}( {\mathbf {r}})^{\infty }$ , then by Lemma 2.5(ii), it follows that as $\beta $ decreases to $\beta _{0}$ , the sequence $\delta (\beta )$ converges to $\mathbb {L}( {\mathbf {r}})^{+}0^{\infty }$ with respect to the order topology, that is, $\lim _{\beta \searrow \beta _{0}}\delta (\beta )=\mathbb {L}( {\mathbf {r}})^{+}0^{\infty }$ . Since $\Psi _ {\mathbf {S}}$ is increasing, it follows that

$$ \begin{align*} \Psi_ {\mathbf{S}}((1,2])=(J^ {\mathbf{S}}\backslash I^ {\mathbf{S}})\backslash \bigcup_{ {\mathbf{r}}\in\Omega_{L}^{*}} (p^{ {\mathbf{r}}},q^{ {\mathbf{r}}}], \end{align*} $$

where $\delta (p^ {\mathbf {r}})=\Phi _ {\mathbf {S}}(\mathbb {L}( {\mathbf {r}})^{\infty })$ and $\delta (q^ {\mathbf {r}})=\Phi _ {\mathbf {S}}(\mathbb {L}( {\mathbf {r}})^{+}0^{\infty })$ . Note that $(p^ {\mathbf {r}},q^ {\mathbf {r}}]\subset J^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ . By Lemma 2.9(iii), there is a (unique) Farey word $\hat { {\mathbf {r}}}$ such that $J^ {\mathbf {r}}\subset J^{\hat r}$ . Applying equation (5.6) to both $ {\mathbf {r}}$ and $\hat { {\mathbf {r}}}$ , and using Lemma 5.3, we conclude that $J^{ {\mathbf {S}}\bullet {\mathbf {r}}}\subset J^{ {\mathbf {S}}\bullet \hat { {\mathbf {r}}}}$ . Hence, $(p^ {\mathbf {r}},q^ {\mathbf {r}}]\subset J^{ {\mathbf {S}}\bullet \hat { {\mathbf {r}}}}$ . Therefore, if $\beta \in E^ {\mathbf {S}}=(J^ {\mathbf {S}}\backslash I^ {\mathbf {S}})\backslash \bigcup _{ {\mathbf {r}}\in \Omega _{F}^{*}} J^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ , then $\beta $ lies in the range of $\Psi _ {\mathbf {S}}$ . This implies that

(5.7) $$ \begin{align} E^ {\mathbf{S}}\subset\Psi_ {\mathbf{S}}((1,2]). \end{align} $$

Now assume first that $\beta \in \Psi _ {\mathbf {S}}(E)$ . Then $\beta =\Psi _ {\mathbf {S}}({\hat \beta })$ , where ${\hat \beta }\not \in J^ {\mathbf {r}}$ for any $ {\mathbf {r}}\in \Omega _{F}^{*}$ . Hence, $\beta \not \in J^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ for any $ {\mathbf {r}}\in \Omega _{F}^{*}$ by equation (5.6) and since $\Psi _ {\mathbf {S}}$ is increasing. Therefore, $\beta \in E^ {\mathbf {S}}$ .

Conversely, suppose $\beta \in E^ {\mathbf {S}}$ . By equation (5.7), $\beta =\Psi _ {\mathbf {S}}({\hat \beta })$ for some ${\hat \beta }\in (1,2]$ . If ${\hat \beta }\in J^ {\mathbf {r}}$ for some $ {\mathbf {r}}\in \Omega _{F}^{*}$ , then $\beta \in \Psi _ {\mathbf {S}}(J^ {\mathbf {r}})\subset J^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ by equation (5.6), contradicting that $\beta \in E^ {\mathbf {S}}$ . Hence, ${\hat \beta }\in E$ and then $\beta \in \Psi _ {\mathbf {S}}(E)$ . This completes the proof.

Proof. Since $\delta (p)=\mathbb {L}( {\mathbf {S}})^{\infty }$ and $\delta (q)=\mathbb {L}( {\mathbf {S}})^{+} {\mathbf {S}}^{\infty }$ , we have

$$ \begin{align*} (\mathbb{L}( {\mathbf{S}})^{+}0^{\infty})_{p}=1=(\mathbb{L}( {\mathbf{S}})^{+} {\mathbf{S}}^{\infty})_{q}=:((c_{i}))_{q}. \end{align*} $$

It follows that

$$ \begin{align*} |J^ {\mathbf{S}}|=q-p=\sum_{i=1}^{| {\mathbf{S}}|}\frac{c_{i}}{q^{i-1}}+\sum_{i=| {\mathbf{S}}|+1}^{\infty}\frac{c_{i}}{q^{i-1}}-\sum_{i=1}^{| {\mathbf{S}}|}\frac{c_{i}}{p^{i-1}} \le\sum_{i=| {\mathbf{S}}|+1}^{\infty}\frac{1}{q^{i-1}}=\frac{q}{q-1}q^{-| {\mathbf{S}}|}, \end{align*} $$

as required.

Proof. (i) First we prove $\dim _{P} E=0$ . Let $\rho _{N}\in (1,2]$ such that $\delta (\rho _{N})=(10^{N-1})^{\infty }$ . Then by Lemma 2.5, it follows that $\rho _{N}\searrow 1$ as $N\to \infty $ . Thus $E=\bigcup _{N=1}^{\infty }(E\cap [\rho _{N},2])$ . By the countable stability of packing dimension (cf. [Reference Falconer16]), it suffices to prove that

(5.8) $$ \begin{align} \dim_{B}(E\cap[\rho_{N},2])=0\quad\textrm{for all } N\in\mathbb{N}. \end{align} $$

Let $N\in \mathbb {N}$ . Take a Farey interval $J^ {\mathbf {s}}:=[p, q]\subset [\rho _{N}, 2]$ with $ {\mathbf {s}}=s_{1}\ldots s_{m}\in \Omega _{F}^{*}$ such that

(5.9) $$ \begin{align} m> N+2-3\log_{2}(\rho_{N}-1). \end{align} $$

Write $\mathbb {L}( {\mathbf {s}})=a_{1}\ldots a_{m}$ . Then $\delta (p)=(a_{1}\ldots a_{m})^{\infty }$ . Since $p\ge \rho _{N}$ , by Lemma 2.5, we have $(a_{1}\ldots a_{m})^{\infty }=\delta (p)\succcurlyeq \delta (\rho _{N})=(10^{N-1})^{\infty }$ , which implies that $a_{1}\ldots a_{N+1}\succcurlyeq 10^{N-1}1$ . Then by Proposition 2.4, we conclude that

(5.10) $$ \begin{align} s_{1}\ldots s_{N+1}\succcurlyeq 0^{N}1. \end{align} $$

Note that

$$ \begin{align*} (\mathbb{L}( {\mathbf{s}})^{+} 0^{\infty})_{p}=1=(\mathbb{L}( {\mathbf{s}})^{+}{ {\mathbf{s}}^{\infty}})_{q}=:((c_{i}))_{q}. \end{align*} $$

So, by equation (5.10), it follows that

$$ \begin{align*} \sum_{i=1}^{m}\frac{c_{i}}{p^{i}}=1=\sum_{i=1}^{m}\frac{c_{i}}{q^{i}}+\sum_{i=m+1}^{\infty}\frac{c_{i}}{q^{i}}>\sum_{i=1}^{m} \frac{c_{i}}{q^{i}}+\frac{1}{q^{m+N+1}}, \end{align*} $$

which implies

$$ \begin{align*} \frac{1}{q^{m+N+1}}<\sum_{i=1}^{\infty}\bigg(\frac{1}{p^{i}}-\frac{1}{q^{i}}\bigg)=\frac{q-p}{(p-1)(q-1)}. \end{align*} $$

Whence,

(5.11) $$ \begin{align} |J^ {\mathbf{s}}|=q-p>\frac{(p-1)(q-1)}{q^{N+1}}q^{-m}\ge \frac{(\rho_{N}-1)^{2}}{2^{N+1}}q^{-m}. \end{align} $$

However, by Lemma 5.5, it follows that

(5.12) $$ \begin{align} |J^ {\mathbf{s}}|\le \frac{q}{q-1}q^{-m}\le\frac{2}{\rho_{N}-1}q^{-m} {\leq \frac{2}{\rho_{N}-1}\rho_{N}^{-m}}. \end{align} $$

Now we list all of the Farey intervals in $[\rho _{N},2]$ in a decreasing order according to their length, say $J^{ {\mathbf {s}}_{1}}, J^{ {\mathbf {s}}_{2}}, \ldots $ . In other words, $|J^{ {\mathbf {s}}_{i}}|\ge |J^{ {\mathbf {s}}_{j}}|$ for any $i<j$ . For a Farey interval $J^ {\mathbf {s}}$ , if $J^ {\mathbf {s}}=J^{ {\mathbf {s}}_{k}}$ , we then define its order index as $o(J^ {\mathbf {s}})=k$ .

Set $C_{N}:=2\log 2/\log \rho _{N}$ . Let $J^{ {\mathbf {s}}^{\prime }}$ be a Farey interval with $| {\mathbf {s}}^{\prime }|>C_{N} m$ . Then by equations (5.9), (5.11), and (5.12), it follows that

$$ \begin{align*} |J^{ {\mathbf{s}}^{\prime}}|\le\frac{2}{\rho_{N}-1}\rho_{N}^{-C_{N} m}=\frac{2}{\rho_{N}-1}2^{-2m}<\frac{(\rho_{N}-1)^{2}}{2^{N+1}}2^{-m}\le |J^ {\mathbf{s}}|. \end{align*} $$

This implies that

(5.13) $$ \begin{align} o(J^ {\mathbf{s}})\le\sum_{k=1}^{\lfloor C_{N} m\rfloor}\#\{ {\mathbf{s}}^{\prime}\in\Omega_{F}^{*}: | {\mathbf{s}}^{\prime}|=k\}\le\sum_{k=1}^{\lfloor C_{N} m\rfloor}(k-1)< C_{N}^{2} m^{2}, \end{align} $$

where the second inequality follows by equation (2.1) since the number of non-degenerate Farey words of length k is at most $k-1$ (see [Reference Carminati, Isola and Tiozzo8, Proposition 2.3]). Together with equation (5.12), equation (5.13) implies that

$$ \begin{align*} \liminf_{i\to\infty}\frac{-\log|J^{ {\mathbf{s}}_{i}}|}{o(J^{ {\mathbf{s}}_{i}})}&=+\infty. \end{align*} $$

Note that $[\rho _{N},2]\cap E=[\rho _{N},2]\setminus \bigcup _{ {\mathbf {s}}\in \Omega _{F}^{*}}J^ {\mathbf {s}}$ . So, by [Reference Falconer17, Proposition 3.6], we conclude equation (5.8). This proves $\dim _{P} E=0$ .

Next we prove that E is uncountable. For $ {\mathbf {s}}\in \Omega _{F}^{*}$ , let $\hat J^ {\mathbf {s}}=(\beta _{\ell }^ {\mathbf {s}}, \beta _{r}^ {\mathbf {s}})$ be the interior of the Farey interval $J^ {\mathbf {s}}=[\beta _{\ell }^ {\mathbf {s}},\beta _{r}^ {\mathbf {s}}]$ . By Lemma 2.9(i), it follows that the compact set

$$ \begin{align*} \hat E:=[1,2]\setminus\bigcup_{ {\mathbf{s}}\in\Omega_{F}^{*}}\hat J^ {\mathbf{s}} \end{align*} $$

is non-empty and has no isolated points. Hence, $\hat E$ is a perfect set and is therefore uncountable. Since $\hat {E}\backslash E$ is countable, it follows that E is uncountable as well.

(ii) In a similar way, we prove $\dim _{B} E^ {\mathbf {S}}=0$ . Note that $E^ {\mathbf {S}}=(\beta _{*}^ {\mathbf {S}}, \beta _{r}^ {\mathbf {S}}]\setminus \bigcup _{ {\mathbf {r}}\in \Omega _{F}^{*}}J^{S\bullet {\mathbf {r}}}$ . Fix a Farey word $ {\mathbf {r}}=r_{1}\ldots r_{m}$ . Then the Lyndon interval $J^{ {\mathbf {S}}\bullet {\mathbf {r}}}=[\beta _{\ell }^{ {\mathbf {S}}\bullet {\mathbf {r}}}, \beta _{r}^{ {\mathbf {S}}\bullet {\mathbf {r}}}]=:[p_ {\mathbf {r}}, q_ {\mathbf {r}}]$ satisfies

$$ \begin{align*} {(\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{+} 0^{\infty})_{p_ {\mathbf{r}}}=1=(\mathbb{L}( {\mathbf{S}}\bullet {\mathbf{r}})^{+} \; ( {\mathbf{S}}\bullet {\mathbf{r}})^{\infty})_{q_ {\mathbf{r}}}=:((d_{i}))_{q_ {\mathbf{r}}}.} \end{align*} $$

So,

$$ \begin{align*} \sum_{i=1}^{m| {\mathbf{S}}|}\frac{d_{i}}{p_ {\mathbf{r}}^{i}}=1=\sum_{i=1}^{m| {\mathbf{S}}|}\frac{d_{i}}{q_ {\mathbf{r}}^{i}}+\sum_{i=m| {\mathbf{S}}|+1}^{\infty}\frac{d_{i}}{q_ {\mathbf{r}}^{i}}>\sum_{i=1}^{m| {\mathbf{S}}|}\frac{d_{i}}{q_ {\mathbf{r}}^{i}}+\frac{1}{q_ {\mathbf{r}}^{(m+1)| {\mathbf{S}}|+1}}, \end{align*} $$

where the inequality follows by observing that $ {\mathbf {S}}\in \Omega _{L}^{*}$ and thus $ {\mathbf {S}}\bullet {\mathbf {r}}^{\infty }\succcurlyeq 0^{| {\mathbf {S}}|}10^{\infty }$ . Therefore,

$$ \begin{align*} \frac{1}{q_ {\mathbf{r}}^{(m+1)| {\mathbf{S}}|+1}}\le \sum_{i=1}^{\infty}\bigg(\frac{1}{p_ {\mathbf{r}}^{i}}-\frac{1}{q_ {\mathbf{r}}^{i}}\bigg)=\frac{q_ {\mathbf{r}}-p_ {\mathbf{r}}}{(p_ {\mathbf{r}}-1)(q_ {\mathbf{r}}-1)}, \end{align*} $$

which implies

$$ \begin{align*} |J^{ {\mathbf{S}}\bullet {\mathbf{r}}}|=q_ {\mathbf{r}}-p_ {\mathbf{r}}\ge \frac{(p_ {\mathbf{r}}-1)(q_ {\mathbf{r}}-1)}{q_ {\mathbf{r}}^{| {\mathbf{S}}|+1}}q_ {\mathbf{r}}^{-| {\mathbf{S}}\bullet {\mathbf{r}}|}>\frac{(\beta_{*}^ {\mathbf{S}}-1)^{2}}{2^{| {\mathbf{S}}|+1}}q_ {\mathbf{r}}^{-| {\mathbf{S}}\bullet {\mathbf{r}}|}. \end{align*} $$

However, by Lemma 5.5, it follows that

$$ \begin{align*} |J^{ {\mathbf{S}}\bullet {\mathbf{r}}}|\leq \frac{q_ {\mathbf{r}}}{q_ {\mathbf{r}}-1}q_ {\mathbf{r}}^{-m| {\mathbf{S}}|}\le\frac{2}{\beta_{*}^ {\mathbf{S}}-1}q_ {\mathbf{r}}^{-| {\mathbf{S}}\bullet {\mathbf{r}}|}. \end{align*} $$

Now let $(J^{ {\mathbf {S}}\bullet {\mathbf {r}}_{i}})$ be an enumeration of the intervals $J^{ {\mathbf {S}}\bullet {\mathbf {r}}}$ , $ {\mathbf {r}}\in \Omega _{F}^{*}$ , arranged in order by decreasing length. Then by a similar argument as in (i) above, we obtain

$$ \begin{align*} \liminf_{i\to\infty}\frac{-\log|J^{ {\mathbf{S}}\bullet {\mathbf{r}}_{i}}|}{\log o(J^{ {\mathbf{S}}\bullet {\mathbf{r}}_{i}})}{=+\infty}. \end{align*} $$

Thus, $\dim _{B} E^ {\mathbf {S}}=0$ .

Finally, since we showed in (i) that E is uncountable, we conclude by Lemma 5.3 and Proposition 5.4 that $E^ {\mathbf {S}}=\Psi _ {\mathbf {S}}(E)$ is also uncountable. This completes the proof.

Proof. Note by equation (1.7) that

(5.14) $$ \begin{align} E_{\infty}=\bigcap_{n=1}^{\infty}\bigcup_{ {\mathbf{S}}\in\Lambda(n)}J^ {\mathbf{S}} \subset \bigcap_{n=1}^{\infty}\bigcup_{ {\mathbf{S}}\in\Lambda:| {\mathbf{S}}|\geq n}J^ {\mathbf{S}}. \end{align} $$

This suggests covering $E_{\infty }$ by the intervals $J^ {\mathbf {S}}$ for $ {\mathbf {S}}\in \Lambda $ with $| {\mathbf {S}}|\geq n$ for a sufficiently large n. To this end, we first estimate the diameter of $J^ {\mathbf {S}}$ . Take $ {\mathbf {S}}\in \Lambda $ with $| {\mathbf {S}}|=m$ , and write $J^ {\mathbf {S}}=[p,q]$ . Then $\delta (p)=\mathbb {L}( {\mathbf {S}})^{\infty }$ and $\delta (q)=\mathbb {L}( {\mathbf {S}})^{+} {\mathbf {S}}^{\infty }$ , and it follows from Lemma 5.5 that

(5.15) $$ \begin{align} {|J^ {\mathbf{S}}|\leq \frac{q}{q-1}\cdot q^{-m}.} \end{align} $$

Let $(\beta _{n})$ be an arbitrary sequence in $(1,2)$ decreasing to $1$ . We will use equation (5.15) to show that $\dim _{H} (E_{\infty }\cap (\beta _{n},2])=0$ for all $n\in \mathbb {N}$ , so the result will follow from the countable stability of Hausdorff dimension (cf. [Reference Falconer16]). Fix $n\in \mathbb {N}$ . Observe that if $J^ {\mathbf {S}}=[p,q]$ intersects $(\beta _{n},2]$ , then $q>\beta _{n}$ and so by equation (5.15),