Proof. The first assertion is obvious. For the second assertion,

$$ \begin{align*} \omega^k f_\alpha(\omega^k z)=\omega^{k}(\omega^{2k}z^2+\omega^{-k} \alpha\overline{z})=f_\alpha(z), \end{align*} $$

and hence,

$$ \begin{align*} \omega^k f_\alpha^2(\omega^{-k} z)=\omega^k f_\alpha(\omega^k(\omega^{-k} f_\alpha(\omega^{-k} z)))= f_\alpha(\omega^{-k} f_\alpha(\omega^{-k} z))=f_\alpha^2(z). \\[-37pt] \end{align*} $$

Proof. For simplicity, we only need to prove the case where $k=0$ , since the other cases can be derived by Lemma 2.1. To see that $f_{\alpha }:\Xi _0^0\rightarrow \Delta _0$ and $f_{\alpha }:\Xi _0^1\rightarrow \Delta _0$ are homeomorphisms, we first divide $\Delta _0$ into two parts by $\mathcal {V}$ , and accordingly divide $\Xi _0^0$ and $\Xi _0^1$ into two parts by $\mathcal {G}$ , respectively. Note that for any $i=0,1,2$ ,

$$ \begin{align*} f_\alpha:\omega^i(p_0,+\infty)\rightarrow \omega^{2i}(q_0,+\infty)\quad \text{and}\quad f_\alpha:\omega^i(-\infty,-3p_0)\rightarrow \omega^{2i}(q_0,+\infty) \end{align*} $$

are homeomorphisms. Combining with the mapping properties of $f_\alpha $ mentioned above, it is easy to see that

$$ \begin{align*}f_\alpha:\overline{D_0^0}\rightarrow \overline{V_0}\quad \text{and} \quad f_\alpha:\overline{D_0^1}\rightarrow \overline{V_0}\end{align*} $$

are homeomorphisms, and

$$ \begin{align*}f_\alpha:\Xi_0^0\setminus \overline{D_0^0}\rightarrow \Delta_0\setminus \overline{V_0}\quad \text{and}\quad f_\alpha:\Xi_0^1\setminus\overline{ D_0^1}\rightarrow \Delta_0\setminus \overline{V_0}\end{align*} $$

are homeomorphisms. This finishes the proof of the lemma.

Proof. This follows from Lemma 2.4 and the argument mentioned above.

Proof of Theorem 1.2

Assume that the Julia set $J(f_\alpha )$ is a tree. The proof is by contradiction and is divided into three cases.

Case 1: $\alpha \in \mathbb {R}\setminus [-2,4]$ . According to Lemma 2.6, it follows that the mapping $f_\alpha |_{\mathbb {R}}$ is conjugate to $x \mapsto x^{2}+c$ with $c \in (-\infty ,-2]$ . Hence, $K(f_{\alpha })\cap \mathbb {R}$ is a Cantor set. Since $J(f_\alpha )$ is a tree, there is a Jordan arc $\gamma $ contained in $J(f_\alpha )$ connecting two distinct points in $K(f_{\alpha })\cap \mathbb {R}$ . However, Lemma 2.1 implies that the symmetric image of $\gamma $ about the real axis $\gamma ^*$ is also contained in $J(f_\alpha )$ . Since $K(f_{\alpha })\cap \mathbb {R}$ is a Cantor set, we have $\gamma \neq \gamma ^*$ . For otherwise, $\gamma =\gamma ^*$ would be a straight segment contained in $K(f_{\alpha })\cap \mathbb {R}$ , which is impossible. Since $\gamma \neq \gamma ^*$ , $\gamma \cup \gamma ^*$ must contain a subset homeomorphic to a circle. This is a contradiction.

Case 2: $\alpha \in [-2,-1]$ . Note that $|\alpha |\leq |\sqrt {-2\alpha }|$ . From (2.4) and the last assertion of Lemma 2.2, we get

$$ \begin{align*}\mathrm{Int}(\mathcal{V}_{\alpha})\subseteq\mathrm{Int}(\mathcal{V}_{\sqrt{-2\alpha}})=\mathrm{ Int}(\mathcal{G}_{\alpha}).\end{align*} $$

Since $\mathrm {Int}(\mathcal {V}_{\alpha })=f_{\alpha }(\mathrm {Int}(\mathcal {G}_{\alpha }))$ by Lemma 2.3, we have $\mathrm {Int}(\mathcal {G}_{\alpha }) \subseteq K(f_{\alpha })$ . Hence, $K(f_{\alpha })$ has interior points. This, together with the fact that $\partial K(f_{\alpha })\subseteq J(f_{\alpha })$ and $J(f_{\alpha })$ is locally connected, implies that $\partial K(f_{\alpha })$ contains a subset homeomorphic to a circle. This is a contradiction.

Case 3: $\alpha \in (-1,1)$ . In this case, since the origin is an attracting fixed point, $K(f_{\alpha })$ has interior points. The same argument as in Case 2 will lead to a contradiction.

Proof. We first prove the first assertion with $k=0$ . Suppose $z_0=r e^{\mathrm{i}\theta }\in \overline {U(0,\alpha )}\cap \Delta _0^\sharp $ . Then,

$$ \begin{align*}\arg(f_\alpha(z_0))=\arg(r^2 e^{2\mathrm{i}\theta}+\alpha re^{-\mathrm{i}\theta}).\end{align*} $$

It follows that for $0<\theta <\pi /3$ ,

$$ \begin{align*}-\theta<\arg(f_\alpha(z_0))\leq \arg(\alpha re^{2\mathrm{i}\theta}+\alpha re^{-\mathrm{i}\theta})=\theta/2,\end{align*} $$

and similarly for $-\pi /3<\theta <0$ ,

$$ \begin{align*}\theta/2\leq\arg(f_\alpha(z_0))< -\theta.\end{align*} $$

This proves that

(3.1) $$ \begin{align} |\!\arg(f_\alpha(z_0))|<|\!\arg(z_0)|. \end{align} $$

Let $k\in \{0,1,2\}$ . For any point $z_k\in \overline {U(0,\alpha )}\cap \Delta _k^\sharp $ , we can assume $z_k=\omega ^k z_0$ for ${\Delta _k^\sharp =\omega ^k\Delta _0^\sharp }$ . Then, according to Lemma 2.1,

$$ \begin{align*}f_\alpha(z_k)=f_\alpha(\omega^k z_0)=\omega^{-k}f_\alpha(z_0).\end{align*} $$

This implies that

$$ \begin{align*}|\!\arg(f_\alpha(z_k))+2k\pi/3|=|\!\arg(\omega^{-k}f_\alpha(z_0))+2k\pi/3|=|\!\arg(f_\alpha(z_0))|.\end{align*} $$

In addition, we also have

$$ \begin{align*}|\!\arg(z_k)-2k\pi/3|=|\!\arg(\omega^kz_0)-2k\pi/3|=|\!\arg(z_0)|.\end{align*} $$

Therefore, it follows by (3.1) that the first assertion holds.

The second assertion can be derived by the fact that if $z_k\in (Y\cup \omega ^{1/2} Y)\cap \Delta _k^\sharp $ , then $\arg (f_\alpha (z_k))=-\arg (z_k)$ .

Proof. Let $r>0$ and $\theta \in [0,2\pi )$ . For $z=re^{\mathrm {i}\theta }$ , we have $f_\alpha (z)=z^2+\alpha \bar {z}=r^2e^{2\mathrm {i}\theta }+\alpha re^{-\mathrm {i}\theta }$ . Then,

(3.2) $$ \begin{align} |f_\alpha(z)|^2&=(r^2e^{2\mathrm{i}\theta}+\alpha re^{-\mathrm{i}\theta})(r^2e^{-2\mathrm{i}\theta}+\alpha re^{\mathrm{i}\theta})\notag\\ &= r^2(r^2+2\alpha r\cos3\theta+\alpha^2). \end{align} $$

Suppose $z\in A_{2\alpha }$ . By (3.2), $|f_\alpha (z)|^2\geq r^2(r-\alpha )^2$ and, thus,

(3.3) $$ \begin{align} |f_\alpha(z)|\geq r(r-\alpha). \end{align} $$

This implies that $|f_\alpha (z)|>\alpha r\geq r>2\alpha $ , which means

$$ \begin{align*}f_\alpha(A_{2\alpha})\subseteq A_{2\alpha}.\end{align*} $$

Since $r-\alpha>\alpha \geq 1$ , it follows from (3.3) that $|f_\alpha ^n(z)|\rightarrow +\infty $ as $n\rightarrow +\infty $ . That is,

(3.4) $$ \begin{align} A_{2\alpha} \subseteq E(f_\alpha).\end{align} $$

This proves the first assertion of the lemma.

For $k=0,1,2$ , let

$$ \begin{align*}E_{\pi/6}^k=E_{\pi/6}\cap\Delta_k^\sharp.\end{align*} $$

To prove the second assertion of the lemma, it suffices to prove that

(3.5) $$ \begin{align}f_\alpha(E_{\pi/6}^0)\subseteq E_{\pi/6}^0\cup A_{2\alpha}\quad \text{and}\quad E_{\pi/6}^0 \subseteq E(f_\alpha).\end{align} $$

In fact, since $E_{\pi /6}^k=\omega ^k E_{\pi /6}^0$ with $k=1,2$ , it follows from (3.5) and Lemma 2.1 that

$$ \begin{align*}f_\alpha(E_{\pi/6}^k)=\omega^{-k} f_\alpha(E_{\pi/6}^0)\subseteq E_{\pi/6}^{-k}\cup A_{2\alpha}\quad \text{and}\quad E_{\pi/6}^k \subseteq E(f_\alpha).\end{align*} $$

Now, let us prove (3.5). Note that for $z=re^{\mathrm {i}\theta }\in E_{\pi /6}^0$ , according to (3.2) and ${0\leq \cos (3\theta )\leq 1}$ , we can get that

(3.6) $$ \begin{align} |f_\alpha(z)|\geq r\sqrt{r^2+\alpha^2}>\alpha r. \end{align} $$

For an interval $B\subseteq \mathbb {R}^+$ , define

$$ \begin{align*}E(B):=\{z=re^{\mathrm{i}\theta}:z\in E_{\pi/6}^0\; \text{and}\; r\in B\}.\end{align*} $$

Then, we divide $E_{\pi /6}^0$ into three parts: $E(0,\alpha ]$ , $E(\alpha ,\sqrt {3}\alpha )$ , and $E[\sqrt {3}\alpha ,+\infty )$ .

For the case where $z\in E(0,\alpha ]$ , it follows from Lemma 3.1 that

(3.7) $$ \begin{align}f_\alpha(E(0,\alpha])\subseteq E_{\pi/6}^0.\end{align} $$

For the case where $z=re^{\mathrm {i}\theta }\in E(\alpha ,\sqrt {3}\alpha )$ , according to Lemma 2.1, we only need to analyze the case where $0\leq \theta \leq \pi /6$ . Note that by a simple calculation,

(3.8) $$ \begin{align} \tan(\arg(f_\alpha(z)))=\frac{r\sin(2\theta)-\alpha\sin(\theta)}{r\cos(2\theta)+\alpha\cos(\theta)}. \end{align} $$

Thus, the numerator and the denominator are both positive in this case. Since

$$ \begin{align*}\frac{d}{d\theta}(r\sin(2\theta)-\alpha\sin(\theta))=2r\cos(2\theta)-\alpha\cos(\theta) \geq r-\alpha \geq 0,\end{align*} $$

the numerator is increasing with respect to $\theta $ . However, it is obvious that the denominator $r\cos (2\theta )+\alpha \cos (\theta )$ is decreasing with respect to $\theta $ . This implies that $\tan (\arg (f_\alpha (z)))$ , and hence $\arg (f_\alpha (z))$ , is increasing with respect into $\theta $ . Furthermore, by writing (3.8) as

$$ \begin{align*}\tan(\arg(f_\alpha(z)))=\tan(2\theta)-\frac{\alpha\cos(\theta)\tan(2\theta)+\alpha\sin(\theta)}{r\cos(2\theta)+\alpha\cos(\theta)},\end{align*} $$

it is clear that $\arg (f_\alpha (z))$ is also increasing with respect to r in this case. Thus, the infimum of the argument of $f_\alpha (z)$ is $0$ , which is reached at $\theta =0$ , and the maximum argument of $f_\alpha (z)$ is $\pi /6$ , which is reached at $\theta =\pi /6$ and $r=\sqrt {3}\alpha $ . This proves

(3.9) $$ \begin{align}f_\alpha(E(\alpha,\sqrt{3}\alpha))\subseteq E_{\pi/6}^0.\end{align} $$

For the case where $z\in E[\sqrt {3}\alpha ,+\infty )$ , the first inequality of (3.6) implies that ${|f_\alpha (z)|\geq 2\sqrt {3}\alpha ^2}$ . This proves

(3.10) $$ \begin{align}f_\alpha(E[\sqrt{3}\alpha,+\infty))\subseteq A_{2\alpha}.\end{align} $$

Combining with (3.7), (3.9), and (3.10), we finish the proof of the first conclusion of (3.5).

For the second conclusion of (3.5), it follows from (3.10) and (3.4) that

(3.11) $$ \begin{align}E[\sqrt{3}\alpha,+\infty)\subseteq E(f_\alpha).\end{align} $$

To prove $E(0,\sqrt {3}\alpha )\subseteq E(f_\alpha )$ , let $z\in E(0,\sqrt {3}\alpha )$ . Note that $\mathcal {O}(z)\nsubseteq E(0,\sqrt {3}\alpha )$ . For otherwise, we may get a contradiction by (3.6). Let $n\geq 1$ be the first integer such that $f_\alpha ^n(z)\notin E(0,\sqrt {3}\alpha )$ . By (3.7) and (3.9), we have

$$ \begin{align*}f_\alpha^n(z)\in E[\sqrt{3}\alpha,+\infty).\end{align*} $$

Combining with (3.11), we get $z\in E(f_\alpha )$ . This finishes the proof of the second conclusion of (3.5). The proof of the lemma has been completed.

Proof. For the second assertion, we can first prove that for $\theta _0\in (-\pi /3,\pi /3)$ ,

$$ \begin{align*} \arg(f_\alpha(\alpha e^{\mathrm{i}\theta_0}))=\arg(\alpha^2 e^{2\mathrm{i}\theta_0}+\alpha^2 e^{-\mathrm{i}\theta_0})=\theta_0/2.\end{align*} $$

For $\theta _k=2k\pi /3+\theta _0\in ((2k-1)\pi /3,(2k+1)\pi /3)$ , it follows from Lemma 2.1 that

$$ \begin{align*} \arg(f_\alpha(\alpha e^{\mathrm{i}\theta_k}))=\arg(f_\alpha(\alpha e^{\mathrm{i}\theta_0}))-2k\pi/3=\theta_0/2-2k\pi/3=\theta_k/2-k\pi.\end{align*} $$

This proves the second assertion.

For the first assertion, note that for $|\theta |<\pi /3$ , we have $|\!\arg (f_\alpha (\alpha e^{\mathrm{i}\theta }))|<\pi /6$ . It follows from Lemma 2.4 that for $z_0\in Q_0$ ,

(3.13) $$ \begin{align} |\!\arg(f_\alpha(z_0))|<\pi/6. \end{align} $$

Since $Q_k=\omega ^k Q_0$ , we assume $z_k=\omega ^k z_0\in Q_k$ for $k=1,2$ . Combining with Lemma 2.1 and (3.13), we can get

$$ \begin{align*} |\!\arg(f_\alpha(z_k))+2k\pi/3|=|\!\arg(\omega^{-k}f_\alpha(z_0))+2k\pi/3|=|\!\arg(f_\alpha(z_0))|<\pi/6. \end{align*} $$

This proves the first assertion.

Proof. By a similar argument as the proof of Lemma 3.3, it suffices to prove that $f_\alpha (W_0)\subseteq E_{\pi /6}$ . Combining with the second assertion of Lemmas 2.2 and 2.4, we can get that $f_\alpha $ is a homeomorphism on $\mathrm {Int}(W_0)$ . Therefore, we only need to prove that $f_\alpha (\partial W_0)\subseteq E_{\pi /6}$ .

By the mapping properties of $f_\alpha $ , it is easy to check that

$$ \begin{align*}\arg(f_\alpha(z))=0 \quad \text{for } z\in\partial W_0\cap\partial\Omega_0.\end{align*} $$

According to Lemma 2.1, what we are left to prove is that

(3.14) $$ \begin{align} -\pi/6<\arg(f_\alpha(z))<0 \quad \text{for } z\in\partial W_0\setminus \partial\Omega_0 \quad \text{and} \quad \mathrm{Im}(z)>0. \end{align} $$

Suppose $w=\alpha e^{\mathrm{i}\theta }/2\in \mathcal {C}$ with $\theta \in (0,\pi /3)$ . Then,

$$ \begin{align*} f_\alpha^2(w)&=w^4+2\alpha w^2\overline{w}+(\alpha^2+\alpha)\overline{w}^2+\alpha^2 w\\\notag &=\frac{\alpha^3}{16}[(\alpha e^{4\mathrm{i}\theta}+4e^{\mathrm{i}\theta})+4(\alpha+1)(e^{-2\mathrm{i}\theta}+e^{\mathrm{i}\theta})].\notag \end{align*} $$

Note that for $\alpha \in [1,4]$ , we have

$$ \begin{align*}\theta<\arg((\alpha e^{4\mathrm{i}\theta}+4e^{\mathrm{i}\theta}))\leq 5\theta/2 \quad \text{and} \quad \arg(4(\alpha+1)(e^{-2\mathrm{i}\theta}+e^{\mathrm{i}\theta}))=-\theta/2.\end{align*} $$

Therefore,

(3.15) $$ \begin{align}\arg(f_\alpha^2(w))>-\theta/2>-\pi/6.\end{align} $$

Since $f_\alpha ^{-1}(\mathbb {R})=\mathbb {R}\cup \{z:\mathrm {Re}(z)=\alpha /2\}$ , it follows by the second assertion of Lemma 2.2 that

$$ \begin{align*}\mathrm{Im}(f_\alpha^2(w))>0 \quad \text{or}\quad \mathrm{Im}(f_\alpha^2(w))<0 \quad \text{uniformly\, for } f_\alpha(w)\in\partial W_0\setminus \partial\Omega_0.\end{align*} $$

While for $w_0=\alpha e^{\pi \mathrm {i}/6}/2$ , we have

$$ \begin{align*}\mathrm{Im}(f_\alpha^2(w_0))=(-3\sqrt{3}/2+2)\alpha+4-2\sqrt{3}\leq 6-7\sqrt{3}/2<0.\end{align*} $$

This means $\arg (f_\alpha ^2(w_0))<0$ and, hence,

(3.16) $$ \begin{align} \arg(f_\alpha^2(w))<0 \quad \text{for }f_\alpha(w)\in\partial W_0\setminus \partial\Omega_0. \end{align} $$

Combining with (3.15) and (3.16), we get (3.14). Therefore, $f_\alpha (\partial W_0)\subseteq E_{\pi /6}$ and this finishes the proof of the lemma.

Proof. Let $z\in \Omega _0\setminus Y$ . Since $f_\alpha (\widetilde {W})\subseteq \overline {\mathrm {Int}(\mathcal {V})}$ , we can get

(3.17) $$ \begin{align} f_\alpha^2(z)\in \bigg(\bigcup_{i=0}^2 f_\alpha(W_i)\bigg)\cup \overline{\mathrm{Int}(\mathcal{V})}\subseteq \bigg[\bigg(\bigcup_{i=0}^2 f_\alpha(W_i)\bigg)\cup \bigg(\bigcup_{i=0}^2 W_i\bigg)\bigg]\cup \widetilde{W}. \end{align} $$

Note that Lemmas 3.4 and 3.2 imply

(3.18) $$ \begin{align} \bigg(\bigcup_{i=0}^2 f_\alpha(W_i)\bigg)\cup \bigg(\bigcup_{i=0}^2 W_i\bigg)\subseteq E(f_\alpha). \end{align} $$

Therefore, if $f_\alpha ^2(z)\in (\bigcup _{i=0}^2 f_\alpha (W_i))\cup (\bigcup _{i=0}^2 W_i)$ , then $z\in E(f_\alpha )$ . Otherwise, if $f_\alpha ^2(z)\in \widetilde {W}$ , we will prove $z\in E(f_\alpha )$ by the following claim: there is a smallest integer $N>1$ such that

(3.19) $$ \begin{align} f_\alpha^{2N}(z)\notin \widetilde{W}. \end{align} $$

We prove the claim by contradiction. Suppose that

$$ \begin{align*}f_\alpha^{2n}(z)\in \widetilde{W}\subseteq\overline{\Omega_0}\end{align*} $$

for any integer $n>0$ . Then, we have

$$ \begin{align*} f_\alpha^{2n+1}(z)\in f_\alpha(\widetilde{W})\subseteq\overline{\mathrm{Int}(\mathcal{V})}=\bigg(\bigcup_{i=0}^2 W_i\bigg)\cup\widetilde{W}. \end{align*} $$

Combining with (3.18) and $z\notin E(f_\alpha )$ , it follows that

$$ \begin{align*}f_\alpha^{2n+1}(z)\in\widetilde{W}\subseteq\overline{\Omega_0}\end{align*} $$

for all $n{\kern-1pt}\geq{\kern-1pt} 0$ . Hence, $\mathcal {O}(z){\kern-1pt}\subseteq{\kern-1pt} \overline {\Omega _0}$ . However, Lemma 3.1 implies that there is an integer ${n>0}$ such that $f_\alpha ^n(z)\in E_{\pi /6}$ . This means $z\in E(f_\alpha )$ by Lemma 3.2, which is a contradiction.

According to (3.19), we get that $f_\alpha ^{2j}(z)\in \overline {\Omega _0}$ for any $j=0,1,\ldots ,N-1$ . However, since $f_\alpha $ is a two-to-one mapping from Y to $f_\alpha (Y)$ , it follows from the mapping property of $f_\alpha $ that

$$ \begin{align*} f^{-1}_{\alpha}(Y)\cap\overline{\Omega_0}=Y. \end{align*} $$

Therefore,

(3.20) $$ \begin{align} f_\alpha^{2(N-1)}(z)\in \overline{\Omega_0}\setminus Y. \end{align} $$

Note that (3.17) also holds for points in $\overline {\Omega _0}\setminus Y$ . Then, it follows by (3.20) and (3.17) that

$$ \begin{align*} f_\alpha^{2N}(z)=f_\alpha^{2}(f_\alpha^{2(N-1)}(z))\in \bigg[\bigg(\bigcup_{i=0}^2 f_\alpha(W_i)\bigg)\cup \bigg(\bigcup_{i=0}^2 W_i\bigg)\bigg]\cup \widetilde{W}. \end{align*} $$

However, by (3.19) and (3.18), we can get that

$$ \begin{align*} f_\alpha^{2N}(z)\in \bigg(\bigcup_{i=0}^2 f_\alpha(W_i)\bigg)\cup \bigg(\bigcup_{i=0}^2 W_i\bigg)\subseteq E(f_\alpha).\end{align*} $$

That is, $z\in E(f_\alpha )$ . This completes the proof of the lemma.

Proof. According to Lemma 2.6, it follows that $f_\alpha (I) \subseteq I$ . Hence, by Lemma 2.1,

$$ \begin{align*}f_\alpha(\omega^{k} I) = \omega^{-k} f_\alpha(I) \subseteq \omega^{2k} I.\\[-37pt] \end{align*} $$

Proof. According to Lemmas 3.3 and 3.2, we can get $f_\alpha (\bigcup Q_k)\subseteq E_{\pi /6}\subseteq E(f_\alpha )$ . This means $\bigcup Q_k\subseteq E(f_\alpha )$ . Since $(\bigcup Q_k)\cup \{o_0,o_1,o_2\}=\overline {\bigcup Q_k}$ by (3.12), this proves that for any admissible sequence $\underline {s}$ , we have

$$ \begin{align*}(\zeta_{\underline{s}^2}\setminus\{o_0,o_1,o_2\})\cap \Big(\overline{\bigcup Q_k}\Big)=\emptyset.\end{align*} $$

Note that $\mathbb {C}\setminus \overline {\bigcup Q_k}$ has two components $\Omega _0$ and $\mathbb {C}\setminus \overline {U(0,\alpha )}$ . Since $\zeta _{\underline {s}^2}\setminus \{o_0,o_1,o_2\}$ is connected, it follows that

$$ \begin{align*}\text{either } \zeta_{\underline{s}^2}\setminus\{o_0,o_1,o_2\}\subseteq\Omega_0 \quad \text{or}\quad \zeta_{\underline{s}^2}\setminus\{o_0,o_1,o_2\}\subseteq \mathbb{C}\setminus \overline{U(0,\alpha)}.\end{align*} $$

However, by the construction of $T_2$ , $I_{\underline {s}_2}$ must intersect some $D^i_j$ with $i+j\neq 2$ (see Figure 3 for an illustration). Therefore, $I_{\underline {s}_2}\setminus \{o_0,o_1,o_2\}\nsubseteq \Omega _0$ . Hence, $\zeta _{\underline {s}^2}\setminus \{o_0,o_1,o_2\}\subseteq \mathbb {C}\setminus \overline {U(0,\alpha )}$ .

Proof. We prove the lemma by contradiction. Assume that for any $\delta>0$ , there is a point $z_0$ such that

(4.1) $$ \begin{align} z_0\in \zeta_{\underline{s}^3}\cap \overline{U(0,\alpha+\delta)}. \end{align} $$

Since $\zeta _{\underline {s}^3}\subseteq \zeta _{\underline {s}^2}\setminus \{o_0,o_1,o_2\}$ , by Lemma 4.2, we have

(4.2) $$ \begin{align} z_0\in \zeta_{\underline{s}^2}\setminus\{o_0,o_1,o_2\}\subseteq\mathbb{C}\setminus \overline{U(0,\alpha)}. \end{align} $$

Combining with (4.1) and (4.2), it follows that

$$ \begin{align*} z_0\in \overline{U(0,\alpha+\delta)}\setminus \overline{U(0,\alpha)} \subseteq U(\partial U(0,\alpha),2\delta). \end{align*} $$

Then, by the continuity of $f_{\alpha }$ , we get that

(4.3) $$ \begin{align}f_{\alpha}(z_0)\in U(f_{\alpha}(\partial U(0,\alpha)),\varepsilon)\end{align} $$

for some $\varepsilon>0$ small enough depending only on $\delta $ .

Now, we claim that

(4.4) $$ \begin{align} f_{\alpha}(z_0)\in E_{\pi/6}\cup U(0,\alpha). \end{align} $$

Let us get a contradiction by assuming (4.4). If $f_{\alpha }(z_0)\in E_{\pi /6}$ , we get $f_{\alpha }(z_0)\in E(f_{\alpha })$ by Lemma 3.2. However, since $z_0\in \zeta _{\underline {s}^3}$ , this is impossible. If $f_{\alpha }(z_0)\in U(0,\alpha )$ , since $f_{\alpha }(z_0)\in (S\setminus Y)\cup \{o_0,o_1,o_2\}$ , we can get

$$ \begin{align*}U(0,\alpha)\cap((S\setminus Y)\cup\{o_0,o_1,o_2\})\neq\emptyset.\end{align*} $$

This contradicts with Lemma 4.2.

To prove (4.4), it follows by (4.3) that we only need to prove

(4.5) $$ \begin{align} U(f_{\alpha}(\partial U(0,\alpha)),\varepsilon)\subseteq E_{\pi/6}\cup U(0,\alpha)\end{align} $$

for $\varepsilon>0$ small enough. Since $\{o_0,o_1,o_2\}\subseteq f_{\alpha }^{-1}(0)$ , there is an open $\eta $ -neighborhood of $\{o_0,o_1,o_2\}$ , say V, which is mapped into $U(0,\alpha )$ . However, it follows by Lemma 3.3 that $\partial U(0,\alpha )\setminus V$ is mapped into the interior of $E_{\pi /6}$ . This means there is an open neighborhood W of $\partial U(0,\alpha )\setminus V$ which is mapped into $E_{\pi /6}$ . Then, (4.5) holds since $V\cup W$ contains some neighborhood of $\partial U(0,\alpha )$ .

Proof. By Lemma 4.3, $\zeta _{\underline {s}^3}\subseteq \mathbb {C}\setminus \overline {U(0,\alpha +\delta _0)}$ for any admissible sequence $\underline {s}$ . However, for any $z\in \mathbb {C}\setminus \overline {U(0,\alpha +\delta _0)}$ and $1\leq \alpha \leq 4$ , we have

(4.6) $$ \begin{align} |df_\alpha|=|2z\,dz+\alpha \,d\bar{z}|\geq|2z||dz|-\alpha|d\bar{z}|\geq(\alpha+2\delta_0)|dz|\geq(1+2\delta_0)|dz|. \end{align} $$

Note that by setting $\underline {s}=s_1s_2\cdots s_m\cdots $ , we have

(4.7) $$ \begin{align} I_{s_{m-1}s_m}=f_\alpha(I_{s_{m-2}s_{m-1}s_m})=\cdots= f_\alpha^{m-2}(I_{s_1s_2\cdots s_m}).\end{align} $$

Combining with (4.6) and (4.7), this implies that the length of $I_{\underline {s}_{m}}$ tends to zero uniformly at the rate of exponential growth as $m\rightarrow \infty $ . Then, the lemma follows by the definition of $\zeta _{\underline {s}^n}$ .

Proof. Let us first prove that $\overline {S}=S^{\infty }\cup S$ . It is clear that $S^{\infty }\cup S\subseteq \overline {S}$ . To show $\overline {S}\subseteq S^{\infty }\cup S$ , it suffices to prove that for any $z_0\in \overline {S}\setminus S$ , we have $z_0\in S^{\infty }$ . Since S is dense in $\overline {S}$ , we can find a sequence $z_n\in S$ such that $z_n\rightarrow z_0$ . Furthermore, we suppose $z_n\in I_{s_1^ns_2^n\cdots s_{l_n}^n}$ , where $s_1^ns_2^n\cdots s_{l_n}^n$ is an admissible block of length $l_n>0$ . If the sequence $l_n$ is bounded by a constant $N>1$ , then $z_n\in T_N$ for all $n\geq 1$ . Since $T_N$ is compact, it follows that $z_0\in T_N\subseteq S$ , which is a contradiction. Otherwise, by taking a subsequence of $z_n$ if necessary, we may assume $l_n\rightarrow \infty $ monotonically.

Recall that $\bigcup _{k=0}^2\omega ^k\mathbb {R}^+$ separates the plane into three sectors $\Delta _k$ with $I_k\subseteq \Delta _k$ , $k=0,1,2$ . Since $\bigcup _{k=0}^2\omega ^k\mathbb {R}^+\subseteq E(f_\alpha )$ and $z_0\neq 0$ , it follows that, if there is some $k\in \{0,1,2\}$ so that $\{s_1^n\}$ contains infinitely many k, then $z_0$ belongs to the interior of $\Delta _k$ . This implies there is exactly one $k\in \{0,1,2\}$ , denoted as $k_1$ , which appears in $\{s_1^n\}$ infinitely many times. Note that $f_\alpha (z_n)\rightarrow f_\alpha (z_0)$ , $f_\alpha (z_n)\in I_{s_2^n\cdots s_{l_n}^n}$ , and $f_\alpha (z_0)\neq 0$ . Using the same argument as above, there is exactly one $k\in \{0,1,2\}$ , denoted as $k_2$ , which appears in $\{s_2^n\}$ infinitely many times. In this way, we get an admissible sequence $\underline {k}=k_1k_2\cdots $ . By Lemma 4.4, we must have $z_0=\zeta _{\underline {k}}^{\infty }\in S^\infty $ .

Now, let us prove the union is disjoint. The proof is by contradiction. Assume $\zeta _{\underline {s}}^{\infty }\in S$ for some admissible sequence $\underline {s}$ . Then, $\zeta _{\underline {s}}^{\infty }\in I_{t_1\cdots t_n}$ for some admissible block $t_1\cdots t_n$ . This implies that $f_{\alpha }^{n-1}(\zeta _{\underline {s}}^{\infty })\in I_{t_n}$ . However,

$$ \begin{align*}f_{\alpha}^{n-1}(\zeta_{\underline{s}}^{\infty})\in S^{\infty}\subseteq \mathbb{C}\setminus \overline{U(0,\alpha)}\end{align*} $$

by Lemma 4.3. This is a contradiction since $I_{t_n}\cap (\mathbb {C}\setminus \overline {U(0,\alpha )})=\emptyset $ .

Proof. Note that for the definition of $f_{\alpha ,k}^{-1}$ , the choice of the equality is artificial and can be changed at will, since $f_\alpha (\omega ^k\mathbb {R}^+)=\omega ^{2k}\mathbb {R}^+$ for $k=0,1,2$ . This implies that $f_{\alpha ,\underline {s}_{n}}^{-n}(\omega ^k\mathbb {R}^+)$ is well defined. So, we take $\gamma _k=f_{\alpha ,\underline {s}_{n}}^{-n}(\omega ^k\mathbb {R}^+)$ and the lemma holds by Corollary 2.5.

Proof. The proof of the first assertion is by contradiction. Assume there are two distinct admissible sequences $\underline {s}$ and $\underline {t}$ such that $\zeta _{\underline {s}}^{\infty }=\zeta _{\underline {t}}^{\infty }$ . Let $n>0$ be the largest integer such that $\underline {s}_{n-1}=\underline {t}_{n-1}.$ Then, $\zeta _{\underline {s}^{n}}\cup \zeta _{\underline {t}^{n}}\cup \zeta _{\underline {s}}^{\infty }$ is a Jordan curve containing the point $o_{\underline {s}_{n-1}}$ . By Lemma 4.5, this implies that

$$ \begin{align*}\Upsilon=f_{\alpha}^{n-1}(\zeta_{\underline{s}^{n}}\cup \zeta_{\underline{t}^{n}}\cup \zeta_{\underline{s}}^{\infty})\end{align*} $$

is a Jordan curve, which contains the origin and therefore contains two of the $I_0,I_1$ , and $I_2$ . It follows that $\Upsilon $ must intersect $\bigcup _{k=0}^2\omega ^k\mathbb {R}^+\subseteq E(f_\alpha )$ , which is a contradiction.

The proof of the second assertion is also by contradiction. Assume there is a loop ${\Upsilon \subseteq \overline {S}}$ . It is clear that $\Upsilon \cap S^\infty \neq \emptyset $ . We claim that $\Upsilon \cap S^\infty $ contains at least two points. For otherwise, since $\Upsilon $ is a Jordan curve, we would have two distinct admissible sequences $\underline {s}$ and $\underline {t}$ such that $\zeta _{\underline {s}}^{\infty }=\zeta _{\underline {t}}^{\infty }\in \Upsilon \cap S^\infty $ . This contradicts with the first assertion. By the claim, let $\zeta _{\underline {s}}^{\infty }\neq \zeta _{\underline {t}}^{\infty }\in \Upsilon $ and $n>0$ be the largest integer such that $\underline {s}_{n-1}=\underline {t}_{n-1}.$ Then, by Lemma 4.6, $\zeta _{\underline {s}}^{\infty }$ and $\zeta _{\underline {t}}^{\infty }$ are contained in distinct sectors separated by the three escaping rays landing at $o_{\underline {s}_{n-1}}$ . This contradicts with the fact that $\Upsilon $ is a Jordan curve. The second assertion has been proved.

Proof. By Lemma 4.2, we know that $S\cap U(0,\alpha )=Y\setminus \{o_0,o_1,o_2\}$ . This proves the local connectivity of $\overline {S}$ on $Y\setminus \{o_0,o_1,o_2\}$ . For any $z\in (S\setminus Y)\cup \{o_0,o_1,o_2\}$ , we can get that there is a smallest integer $n>0$ , such that $f_{\alpha }^{n}(z)\in Y\setminus \{o_0,o_1,o_2\}$ . Since $\{z,f_\alpha (z),\ldots ,f_{\alpha }^{n-1}(z)\}\subseteq \mathbb {C}\setminus U(0,\alpha )$ and $\mathcal {C}\subseteq \overline {U(0,\alpha /2)}$ , this implies that $f_\alpha ^{n}$ is a homeomorphism in some neighborhood of z. Therefore, this implies the local connectivity of $\overline {S}$ on $(S\setminus Y)\cup \{o_0,o_1,o_2\}$ and hence, on S.

For any admissible sequence $\underline {s}$ , let us prove the local connectivity of $\overline {S}$ at $\zeta _{\underline {s}}^{\infty }\in S^{\infty }$ . From Lemma 4.6, we can obtain a sequence neighborhood of $\zeta _{\underline {s}}^{\infty }$ , say $U_{\underline {s}_{n}}$ , with $ U_{\underline {s}_{n}}\cap \overline {S}=\overline {S_{\underline {s}_{n}}}\setminus \{o_{\underline {s}_{n-1}}\}.$ According to Lemma 4.4, the diameter of $S_{\underline {s}_{n}}$ converges to zero as $n\rightarrow \infty $ . This proves the local connectivity of $\overline {S}$ at $\zeta _{\underline {s}}^{\infty }.$ The proof of the lemma is finished.

Proof. This follows from the construction of S, and Lemmas 4.7 and 4.8.

Proof. We first define the bijection

$$ \begin{align*}\varphi:\Sigma_3^\infty\rightarrow \overline{S}\setminus S,\;\underline{s}\mapsto\zeta_{\underline{s}}^{\infty}.\end{align*} $$

This implies that $f_\alpha \circ \varphi (\underline {s})=\varphi \circ \sigma (\underline {s})$ . Suppose $\underline {s_n}\in \Sigma _3^\infty $ is a sequence with $\underline {s_n}\rightarrow \underline {s}$ . It follows that for any $n>0$ , there is an integer $k(n)>0$ such that $(\underline {s_n})_{k(n)}=\underline {s}_{k(n)}$ and $k(n)\rightarrow \infty $ as $n\rightarrow \infty $ . Moreover, Lemma 4.4 implies that for $\varepsilon>0$ , we can get an integer $N>0$ such that for any $n\geq N$ ,

$$ \begin{align*}d(o_{(\underline{s_n})_{k(n)}},\zeta_{\underline{s_n}}^\infty)<\varepsilon/2 \quad \text{and} \quad d(o_{\underline{s}_{k(n)}},\zeta_{\underline{s}}^\infty)<\varepsilon/2.\end{align*} $$

Hence, for $n\geq N$ ,

$$ \begin{align*} d(\zeta_{\underline{s_n}}^{\infty},\zeta_{\underline{s}}^{\infty}) \leq d(\zeta_{\underline{s_n}}^{\infty},o_{(\underline{s_n})_{k(n)}}) +d(o_{(\underline{s_n})_{k(n)}},o_{\underline{s}_{k(n)}}) +d(o_{\underline{s}_{k(n)}},\zeta_{\underline{s}}^{\infty})<\varepsilon. \end{align*} $$

This proves that $\varphi $ is continuous, so $\varphi $ is a homeomorphism. Therefore, the space $(\overline {S}\setminus S, d)$ is topologically conjugate to the space $(\Sigma _3^\infty , \rho )$ .

Then, we show that the topological entropy of $f_{\alpha }$ on $\overline {S}$ equals to $\ln 2$ . On one hand, since $S = \bigcup _{n=0}^{\infty }f_\alpha ^{-n}(Y)$ , it follows from Lemmas 2.6 and 2.1 that the topological entropy of $f_\alpha $ on S is no more than $\ln 2$ . On the other hand, according to (1.2), we get that the topological entropy of $f_\alpha $ on $\overline {S}\setminus S $ is equal to $\ln 2$ . In all, this proves that the topological entropy of $f_\alpha $ on $\overline {S}$ is $\ln 2$ .

Proof. By the definition of $f^{-n}_{\alpha ,0_{n}}$ , it is obvious that $J_{0_{n}}\subseteq \mathbb {C}\setminus \Omega _0$ for all $n\geq 2$ . That is, $\xi _{\overline {0}}\subseteq \mathbb {C}\setminus \Omega _0$ . Since $J_{0_{n}}$ is path connected by Corollary 2.5, and $p_{0_{n}}$ is the common endpoint of $J_{0_{n}}$ and $J_{0_{n+1}}$ , this proves $\xi _{\overline {0}}$ is path connected. Now, let us prove that $\xi _{\overline {0}}$ is non-self-intersecting. On one hand, by Corollary 2.5, it is obvious that $J_{0_{n}}$ is non-self-intersecting for every $n\geq 2$ . On the other hand, suppose that there are two integers i and j with $i>j>1$ such that $J_{0_{i}}\cap J_{0_{j}}\neq \emptyset $ . Then, we can get ${J_{0_{i-j+2}}\cap J_{0_2}\neq \emptyset }$ . Since $f_\alpha (J_{0_2})\subseteq \mathbb {R}^+$ and $f_\alpha (\mathbb {R}^+)=\mathbb {R}^+$ , it follows that $J_{0_{m}}\cap \mathbb {R}^+\neq \emptyset $ for all $2\leq m\leq i-j+1$ . However, by noting that $J_{0_2}\cap \mathbb {R}^+=\emptyset $ , we can obtain

$$ \begin{align*}J_{0_3}\cap \mathbb{R}^+=\emptyset \quad \text{and} \quad J_{0_{2}}\cap J_{0_{3}}=\emptyset.\end{align*} $$

The first equality implies that we must have $i=j+1$ . Then, combining with the second equality and Corollary 2.5, it follows that $J_{0_{i}}\cap J_{0_{j}}=\emptyset $ , which is a contradiction. This implies $\xi _{\overline {0}}$ is non-self-intersecting.

Proof. Note that

$$ \begin{align*}\overline{U(0,\alpha)}=\Omega_0\cup\Big(\bigcup o_k\Big)\cup\Big(\bigcup Q_k\Big).\end{align*} $$

On one hand, from Lemma 5.1, we know that

$$ \begin{align*}A_{\overline{0}}\cap\Omega_0=\emptyset.\end{align*} $$

Moreover, since $f_{\alpha }(A_{\overline {0}})\subseteq A_{\overline {0}}$ and $f_\alpha (o_k)=0\in \Omega _0$ for $k=0,1,2$ , it follow that

$$ \begin{align*}A_{\overline{0}}\cap\Big(\bigcup o_k\Big)=\emptyset.\end{align*} $$

On the other hand, to prove

(5.1) $$ \begin{align}A_{\overline{0}}\cap\Big(\bigcup Q_k\Big)=\emptyset,\end{align} $$

we first claim that $|z|\leq 2\alpha $ for any $z\in A_{\overline {0}}$ . For otherwise, there must be a point ${w\in \xi _{\overline {0}}}$ such that $|w|> 2\alpha $ . According to the first assertion of Lemma 3.2, it follows that $|f_\alpha ^{n}(w)|> 2\alpha $ for all $n\geq 0$ . While from $w\in \xi _{\overline {0}}$ , we can also obtain that there is an integer $m\geq 0$ such that $f_\alpha ^m(w)\in J_{0_2}$ . This is a contradiction. Therefore, $A_{\overline {0}}$ is compact in the complex plane. Moreover, since $f_{\alpha }(A_{\overline {0}})\subseteq A_{\overline {0}}$ , this yields $A_{\overline {0}}\cap E(f_{\alpha })=\emptyset .$ However, combining with Lemmas 3.3 and 3.2, we know that $\cup Q_k\subseteq E(f_{\alpha })$ . Therefore, we get (5.1) holds. In all, we have obtained that

$$ \begin{align*}A_{\overline{0}}\cap\overline{U(0,\alpha)}=\emptyset.\end{align*} $$

Since the set $A_{\overline {0}}$ is closed, it follows that there is a $\delta _1>0$ such that $A_{\overline {0}}$ is contained in $\mathbb {C}\setminus \overline {U(0,\alpha +2\delta _1)}$ . This proves that $\xi _{\overline {0}}(t)$ for every $t>0$ small enough is contained in $\mathbb {C}\setminus \overline {U(0,\alpha +\delta _1)},$ so is $J_{0_m}$ for every integer $m\geq K$ with $K>0$ large enough. The subsequent proof is similar with Lemma 4.4. Note that for any $z\in \mathbb {C}\setminus \overline {U(0,\alpha +\delta _1)}$ and $1\leq \alpha \leq 4$ , we can get that

$$ \begin{align*} |df_\alpha|=|2z\,dz+\alpha \,d\bar{z}|\geq|2z||dz|-\alpha|d\bar{z}|\geq(\alpha+2\delta_1)|dz|\geq(1+2\delta_1)|dz|. \end{align*} $$

It follows by $f_{\alpha }^{m-K}(J_{0_{m}})=J_{0_{K}}$ that the Euclidean length of $J_{0_{m}}$ goes towards zero at the rate of exponential growth as $m\rightarrow \infty $ . Then, the Euclidean length of $\overline {\bigcup _{n\geq m}J_{0_{n}}}$ goes towards zero as $m\rightarrow \infty $ . Hence, the set

$$ \begin{align*} A_{\overline{0}}=\mathop{\bigcap}\limits_{t>0}\overline{\xi_{\overline{0}}((0,t))} =\mathop{\bigcap}\limits_{m\geq K}\overline{\mathop{\bigcup}\limits_{n\geq m}J_{0_{n}}} \end{align*} $$

is a singleton. That is, $\xi _{\overline {0}}$ is landing.

To prove $A_{\overline {0}}=\zeta _{\overline {0}}^{\infty }$ , we first note that $A_{\overline {0}}$ is a fixed point of $f_\alpha $ . Since $J_{0_2}\subseteq \overline {X_0}\subseteq \overline {Y_0}$ and $f_\alpha (\overline {X_0})=\overline {Y_0}$ by the mapping properties of $f_\alpha $ (see Figure 8), it follows that

$$ \begin{align*} \xi_{\overline{0}}\cup A_{\overline{0}}\subseteq\overline{X_0}.\end{align*} $$

Moreover, we claim that

$$ \begin{align*}(\xi_{\overline{0}}\cup A_{\overline{0}})\cap \mathbb{R}^+=\emptyset.\end{align*} $$

For otherwise, there must be a point $w\in \xi _{\overline {0}}\cup A_{\overline {0}}$ such that $|f_\alpha ^{n+1}(w)|>|f_\alpha ^n(w)|$ for all $n\geq 0$ . However, this will lead to a contradiction since if $w\in \xi _{\overline {0}}$ , we can get that there is an integer $m\geq 0$ such that $f_\alpha ^m(w)\in J_{0_2}$ and if $w\in A_{\overline {0}}$ , we can get that $w=A_{\overline {0}}$ is the fixed point of $f_{\alpha }$ .

However, recall that

$$ \begin{align*}\zeta_{0_n}=\mathop{\bigcup}\limits_{m\geq n} I_{0_m} \quad \text{and} \quad \zeta_{\overline{0}}^\infty=\mathop{\bigcap}\limits_{n\geq 0}\overline{\zeta_{0_n}}.\end{align*} $$

Moreover, it follows from the mapping properties of $f_\alpha $ that $I_{0_2}\subseteq \overline {X_0}$ . Hence, by the same argument as above, we can also get

$$ \begin{align*}\zeta_{0_2}\cup\zeta^{\infty}_{\overline{0}}\in\overline{X_0} \quad \text{and} \quad (\zeta_{0_2}\cup\zeta^{\infty}_{\overline{0}})\cap\mathbb{R}^+=\emptyset.\end{align*} $$

Note that it is easy to check that the only fixed points of $f_{\alpha }$ not lying in $\mathbb {R}$ are $\zeta ^{\infty }_{\overline {0}}$ and $\zeta ^{\infty }_{\overline {2}}$ , which are conjugated to each other about the real axis. Therefore, since both $J_{0_2}$ and $I_{0_2}$ lie in the upper half-plane, it follows that $A_{\overline {0}}=\zeta ^{\infty }_{\overline {0}}$ .

Proof. The strategy of the proof is similar as the proof in Lemma 5.2. Note that

$$ \begin{align*}\overline{U(0,\alpha)}=\Omega_0\cup\Big(\bigcup o_k\Big)\cup\Big(\bigcup Q_k\Big).\end{align*} $$

First, we have known from (5.4) that

(5.5) $$ \begin{align}\overline{\xi_{\underline{s}}}\cap\Omega_0=\emptyset.\end{align} $$

Second, since $f_\alpha (\overline {\xi _{\underline {s}}})=\overline {\xi _{\sigma (\underline {s})}}$ and $f_\alpha (o_k)=0$ for $k=0,1,2$ , it follows by (5.4) again that

(5.6) $$ \begin{align}\overline{\xi_{\underline{s}}}\cap\Big(\bigcup o_k\Big)=\emptyset.\end{align} $$

Finally, to show that

(5.7) $$ \begin{align}\overline{\xi_{\underline{s}}}\cap\Big(\bigcup Q_k\Big)=\emptyset\end{align} $$

holds, we divide the proof into two cases.

Case 1: Suppose $n=1$ and $\underline {s}=s_1\overline {xy}\in \Sigma ^1_3\setminus \Sigma ^{0}_3$ . Since by the mapping properties of $f_\alpha $ we have $J_{0_2}\subseteq \overline {X_0}\subseteq \overline {Y_0}$ and $f_\alpha (\overline {X_0})=\overline {Y_0}$ (see Figure 8), it follows that $\overline {\xi _{\overline {0}}}\subseteq \overline {X_0}$ . Then, according to (5.2), we get

$$ \begin{align*}\overline{\xi_{\overline{xy}}}\subseteq \overline{X_k}\subseteq \overline{Y_l}\end{align*} $$

for some $k,l=0,1,2$ . By noting that $\underline {s}\in \Sigma ^1_3\setminus \Sigma ^{0}_3$ , this implies that $\overline {\xi _{\underline {s}}}\subseteq \overline {X^{\prime }_l}$ . Since ${\overline {X^{\prime }_l}\cap (\bigcup Q_k)=\emptyset }$ , we get that (5.7) holds for the case $n=1$ .

Case 2: Suppose $n>1$ . Then, Lemma 3.3 implies that to prove (5.7), it suffices to prove

(5.8) $$ \begin{align}\overline{\xi_{\sigma(\underline{s})}}\cap E_{\pi/6}=\emptyset.\end{align} $$

Now, let us prove (5.8) by contradiction. Assume there is a point $z_0\in \overline {\xi _{\sigma (\underline {s})}}\cap E_{\pi /6}$ . Then, on one hand, by Lemma 3.2, there is an $l\geq 0$ such that

(5.9) $$ \begin{align}\{z_0,f_\alpha(z_0),\ldots, f_\alpha^l(z_0)\}\subseteq E_{\pi/6} \quad \text{and} \quad \{f_\alpha^{l+1}(z_0),f_\alpha^{l+2}(z_0),\ldots\}\subseteq A_{2\alpha}.\end{align} $$

On the other hand, by Case 1, we also have

$$ \begin{align*}f_\alpha^{n-2}(z_0)\in \overline{\xi_{\sigma^{n-1}(\underline{s})}}\subseteq\overline{X^{\prime}_k}\end{align*} $$

for some $k=0,1,2$ . Since $\overline {X^{\prime }_k}\cap E_{\pi /6}=\emptyset $ , (5.9) implies that

(5.10) $$ \begin{align}\{f_\alpha^{n-2}(z_0),f_\alpha^{n-1}(z_0),\ldots\}\subseteq A_{2\alpha}.\end{align} $$

Note that $f_\alpha ^{n-1}(z_0)\in \overline {\xi _{\sigma ^{n}(\underline {s})}}=\overline {\xi _{\overline {xy}}}$ and $\zeta ^{\infty }_{\overline {xy}}=f^2_{\alpha }(\zeta ^{\infty }_{\overline {xy}})\notin A_{2\alpha }$ by Lemma 3.2. It follows that $f_\alpha ^{n-1}(z_0)\in \xi _{\overline {xy}}.$ By the construction (5.2) and Lemma 2.1, we can get that

$$ \begin{align*}f_\alpha^{n-1+m}(z_0)\in \bigcup_{k=0}^2 \omega^k(\overline{J_{0_2}\cup J_{0_2}^*})\subseteq U(0,\alpha)\end{align*} $$

for some $m\geq 0$ . This contradicts with (5.10). Hence, the proof of (5.8) and (5.7) is finished.

Combining with (5.5)–(5.7), it follows that

$$ \begin{align*}\overline{\xi_{\underline{s}}}\cap\overline{U(0,\alpha)}=\emptyset.\end{align*} $$

Since the set $\overline {\xi _{\underline {s}}}$ is closed, the first assertion of the lemma holds. The second assertion holds by a similar argument as the proof of Lemma 4.4.

Proof. Let $\Gamma ^\infty =\Gamma \setminus (\bigcup \Gamma _n)$ . It suffices to prove that $\Gamma ^\infty =S^\infty .$

Set

$$ \begin{align*}R^{\infty}=\overline{\bigcup f_{\alpha,\underline{s}_n}^{-n}(\zeta^{\infty}_{\overline{xy}})},\end{align*} $$

where the union is taken over all $n\geq 0$ and all $\underline {s}=\underline {s}_n \overline {xy}\in \Sigma ^n_3\setminus \Sigma ^{n-1}_3$ . By Lemma 4.10, we get that

$$ \begin{align*}R^\infty=\{\zeta^{\infty}_{\underline{t}}:\text{ for any } \underline{t}\in\Sigma^\infty_3\}=S^\infty.\end{align*} $$

Now, we prove that $\Gamma ^\infty =R^{\infty }.$ On one hand, for $z\in \Gamma ^{\infty }$ , there is a sequence ${z_n\in \bigcup \Gamma _n}$ such that $z_n\rightarrow z$ . Suppose that $z_n$ belongs to $\xi _{\underline {s_n}}$ for some $\underline {s_n}\in \Sigma ^{l_n}_3\setminus \Sigma ^{l_n-1}_3$ with $l_n>0$ . If there is a subsequence of n, say $n_k$ , such that $l_{n_k}\rightarrow \infty $ as $n\rightarrow \infty $ , then by Lemma 5.3, we get that $z_{n_k}\rightarrow z$ is equivalent to $\zeta ^{\infty }_{\underline {s_{n_k}}}\rightarrow z$ . This means $z\in R^{\infty }$ . Otherwise, if the sequence $l_{n}$ is bounded, then there is a subsequence of n, say $n_k$ , such that $\underline {s_{n_k}}=\underline {s_{n_1}}$ for all $k\geq 1$ . By the definition of $\Gamma ^\infty $ , this implies that $z=\zeta ^{\infty }_{\underline {s_{n_1}}}\in R^{\infty }$ . In all, we get that ${\Gamma ^{\infty }\subseteq R^{\infty }}$ . On the other hand, it is obvious that $R^{\infty }\subseteq \Gamma $ . Since $R^{\infty }\cap E(f_\alpha )=\emptyset $ while $\bigcup \Gamma _n\subseteq E(f_\alpha )$ by Lemma 6.1, it follows that $R^{\infty }\subseteq \Gamma ^{\infty }$ . The proof of the lemma is finished.

Proof. For the first assertion, let $\Sigma ^{n,k}_3$ , $k=0,1,2$ , be the subset of $\Sigma ^n_3$ with $s_1=k$ . Let $\underline {s}_{\min }^k(n)\in \Sigma ^{n,k}_3$ be the element such that $\underline {s}_{\min }^k(n)\lhd \underline {s}(n)$ for any $\underline {s}(n)\in \Sigma ^{n,k}_3$ and $\underline {s}_{\max }^k(n)\in \Sigma ^{n,k}_3$ be the element such that $\underline {s}(n)\lhd \underline {s}_{\max }^k(n)$ for any $\underline {s}(n)\in \Sigma ^{n,k}_3$ . Since $0\leq l\leq n$ , we claim that

(5.11) $$ \begin{align} \underline{s}^{\prime}(n)=s_1^{\prime}s_2^{\prime}\cdots s_{l}^{\prime}\underline{s}^{s_{l+1}^{\prime}}_{\max}(n-l)\quad \text{and} \quad\underline{s}^{\prime\prime}(n)=s_1^{\prime\prime}s_2^{\prime\prime}\cdots s_{l}^{\prime\prime}\underline{s}^{s_{l+1}^{\prime\prime}}_{\min}(n-l). \end{align} $$

In fact, only if the form of $\underline {s}^{\prime }(n)$ and $\underline {s}^{\prime \prime }(n)$ are defined by (5.11), we have $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^n_3}$ (this can be checked easily by contradiction; we leave the proof to the readers). However, for any $m\geq 0$ ,

$$ \begin{align*}\underline{s}_{\max}^0(m)=\overline{01},\quad\underline{s}_{\max}^1(m)=\overline{10},\quad\underline{s}_{\max}^2(m)=\overline{2}\end{align*} $$

and

$$ \begin{align*}\underline{s}_{\min}^0(m)=\overline{0},\quad\underline{s}_{\min}^1(m)=\overline{12},\quad\underline{s}_{\min}^2(m)=\overline{21}.\end{align*} $$

This, combined with (5.11), proves $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^l_3$ .

Then, we divide the proof of the first assertion into two cases.

Case 1: Suppose $l=0$ . Then, $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^0_3$ . Since $\Sigma ^0_3=\{\overline {0},\overline {12},\overline {21},\overline {2},\overline {01},\overline {10}\}$ , the relationship $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^n_3}$ is equivalent to:

1. (a) $\underline {s^{\prime }}(n)=\overline {01}$ and $\underline {s^{\prime \prime }}(n)=\overline {12}$ ; or

2. (b) $\underline {s^{\prime }}(n)=\overline {10}$ and $\underline {s^{\prime }}(n)=\overline {21}$ .

In both cases, these imply that $x^{\prime }\equiv x^{\prime \prime }-1\;(\text {mod } 3)$ , $y^{\prime }\equiv y^{\prime \prime }-1\;(\text {mod } 3)$ , and that $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^\infty _3}$ .

Case 2: Suppose $l\in [1,n]$ . For the rest of the proof, we need the following claim:

(5.12) $$ \begin{align} \text{if } l>0,\text{ then } \underline{s^{\prime}}(n)\lhd\underline{s^{\prime\prime}}(n)\text{ is equivalent to } s_{l+1}^{\prime}\equiv s_{l+1}^{\prime\prime}-1\;(\text{mod } 3). \end{align} $$

In fact, $\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)$ only consists of the following three cases: if $s_l^{\prime }=s_l^{\prime \prime }=0$ , then $s_{l+1}^{\prime }=0$ and $s_{l+1}^{\prime \prime }=1$ ; if $s_l^{\prime }=s_l^{\prime \prime }=1$ , then $s_{l+1}^{\prime }=2$ and $s_{l+1}^{\prime \prime }=0$ ; otherwise, if $s_l^{\prime }=s_l^{\prime \prime }=2$ , then $s_{l+1}^{\prime }=1$ and $s_{l+1}^{\prime \prime }=2$ . In all, we have $s_{l+1}^{\prime }\equiv s_{l+1}^{\prime \prime }-1\;(\text {mod } 3)$ .

Since $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^l_3}$ , it follows from (5.12) that the following hold.

1. (a) If $s^{\prime }_{l}=0$ , then $s^{\prime }_{l+1}=0$ and $s^{\prime \prime }_{l+1}=1$ . Therefore, $\underline {s}^{s_{l+1}^{\prime }}_{\max }(n-l)=\overline {01}$ and $\underline {s}^{s_{l+1}^{\prime \prime }}_{\min } (n-l)=\overline {12}$ .

2. (b) If $s^{\prime }_{l}{\kern-1pt}={\kern-1pt}1$ , then $s^{\prime }_{l+1}{\kern-1pt}={\kern-1pt}2$ and $s^{\prime \prime }_{l+1}{\kern-1pt}={\kern-1pt}0$ . Therefore, $\underline {s}^{s_{l+1}^{\prime }}_{\max }(n{\kern-1pt}-{\kern-1pt}l){\kern-1pt}={\kern-1pt}\overline {2}$ and $\underline {s}^{s_{l+1}^{\prime \prime }}_{\min }(n{\kern-1pt}-{\kern-1pt}l)=\overline {0}$ .

3. (c) If $s^{\prime }_{l}=2$ , then $s^{\prime }_{l+1}=1$ and $s^{\prime \prime }_{l+1}=2$ . Therefore, $\underline {s}^{s_{l+1}^{\prime }}_{\max }(n-l)=\overline {10}$ and $\underline {s}^{s_{l+1}^{\prime \prime }}_{\min } (n-l)=\overline {21}$ .

In all cases, these imply that $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^l_3\setminus \Sigma ^{l-1}_3$ , and moreover, $x^{\prime }\equiv x^{\prime \prime }-1\;(\text {mod } 3)$ and $y^{\prime }\equiv y^{\prime \prime }-1\;(\text {mod } 3)$ . In addition, since we have shown above that $\underline {s}_{\max }^k(m)$ and $\underline {s}_{\min }^k(m)$ are independent of m for any $m\geq 0$ , it follows that $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^\infty _3}$ .

For the second assertion, $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^{0}_3$ is obvious if $l=1$ . For the case where $l>1$ , let us prove that either $\underline {s^{\prime }}(n)\in \Sigma ^{n-1}_3$ , $\underline {s^{\prime \prime }}(n)\in \Sigma ^n_3\setminus \Sigma ^{n-1}_3$ or $\underline {s^{\prime }}(n)\in \Sigma ^n_3\setminus \Sigma ^{n-1}_3$ , $\underline {s^{\prime \prime }}(n)\in \Sigma ^{n-1}_3$ by contradiction. On one hand, we assume that $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^{n-1}_3$ . Since $l=n+1$ , this implies that $\underline {s^{\prime }}(n)=\underline {s^{\prime \prime }}(n)$ . This contradicts to $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^n_3}$ . On the other hand, we assume that $\underline {s^{\prime }}(n),\underline {s^{\prime \prime }}(n)\in \Sigma ^{n}_3\setminus \Sigma ^{n-1}_3$ . It follows from $l=n+1$ that $s_{n}^{\prime }=s_{n}^{\prime \prime }$ , $s_{n+1}^{\prime }=s_{n+1}^{\prime \prime }$ , and $s_{n+2}^{\prime }\neq s_{n+2}^{\prime \prime }$ . To get a contradiction, it suffices to prove that $s_n^{\prime }=s_{n+2}^{\prime }$ or $s_n^{\prime \prime }=s_{n+2}^{\prime \prime }$ . In fact, if $s_n^{\prime }\neq s_{n+2}^{\prime }$ and $s_n^{\prime \prime }\neq s_{n+2}^{\prime \prime }$ , then $s_n^{\prime }$ , $s_{n+2}^{\prime }$ , and $s_{n+2}^{\prime \prime }$ will be three different integers. However, $s_{n+1}^{\prime }+s_{n}^{\prime }\neq 2$ , $s_{n+1}^{\prime }+s_{n+2}^{\prime }\neq 2$ , and $s_{n+1}^{\prime }+s_{n+2}^{\prime \prime }\neq 2$ , which is a contradiction.

Moreover, the claim (5.12) shows that $y^{\prime }\equiv y^{\prime \prime }-1\;(\text {mod } 3)$ . In addition, since ${\underline {t^{\prime }}(n{\kern-1pt}+{\kern-1pt}1){\kern-1pt}={\kern-1pt}s_1s_2\cdots s_nx^{\prime }\overline {y^{\prime }z^{\prime }}}$ with $z^{\prime }{\kern-1pt}\equiv{\kern-1pt} x^{\prime }+1\;(\text {mod } 3)$ and $\underline {t^{\prime \prime }}(n+1){\kern-1pt}={\kern-1pt}s_1s_2\cdots s_nx^{\prime \prime }\overline {y^{\prime \prime }z^{\prime \prime }}$ with $z^{\prime \prime }\equiv x^{\prime \prime }-1\;(\text {mod } 3)$ , this implies $\underline {t^{\prime }}(n+1),\underline {t^{\prime \prime }}(n+1)\in \Sigma ^{n+1}_3\setminus \Sigma ^n_3$ . Meanwhile, the claim (5.12) also shows that

$$ \begin{align*}\underline{s^{\prime}}(n)\lhd\underline{t^{\prime}}(n+1)\lhd\underline{t^{\prime\prime}}(n+1)\lhd\underline{s^{\prime\prime}}(n).\end{align*} $$

However, since the first $(n+1)$ numbers of these admissible blocks are the same, there are only four elements in $\Sigma ^{n+1}_3$ with this property. Hence,

$$ \begin{align*}\{\underline{s^{\prime}}(n)\lhd\underline{t^{\prime}}(n+1)\lhd\underline{t^{\prime\prime}}(n+1)\lhd\underline{s^{\prime\prime}}(n)\}_{\Sigma^{n+1}_3}.\\[-39pt] \end{align*} $$

Proof. First of all, we would like to mention that the notation in Lemma 5.5 will be used without extra explanation in this proof.

Part I: divide $\mathbb {S}$ into several pieces inductively to match all the pieces of $\Gamma $ (see Figure 9 for a sketch). Specifically, we will define $\unicode{x3bb} _{\underline {s}}$ for any $\underline {s}\in \bigcup _{k\geq 0}\Sigma ^k_3$ inductively. For the case where $k=0$ , we first define $\unicode{x3bb} _{\underline {s}(0)}$ for any $\underline {s}(0)\in \Sigma ^0_3$ . Set

Similarly to (5.2), we set

Note that $\Sigma ^0_3=\{\overline {0},\overline {12},\overline {21},\overline {2},\overline {01},\overline {10}\}$ and for any $\underline {s^{\prime }}(0), \underline {s^{\prime \prime }}(0)\in \Sigma ^0_3$ with $\{\underline {s^{\prime }}(0)\lhd \underline {s^{\prime \prime }}(0)\}_{\Sigma ^0_3}$ , one of the following statements holds:

• • if $l=0$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(0)}$ and $\unicode{x3bb} _{\underline {s^{\prime \prime }}(0)}$ have a common endpoint;

• • if $l=1$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(0)}\cap \unicode{x3bb} _{\underline {s^{\prime \prime }}(0)}=\emptyset $ .

For the case where $k=n\geq 0$ , suppose that $\unicode{x3bb} _{\underline {s}(n)}$ has been defined for any $\underline {s}(n)\in \Sigma ^n_3$ with the property that for any $\underline {s^{\prime }}(n), \underline {s^{\prime \prime }}(n)\in \Sigma ^n_3$ with $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^n_3}$ , one of the following statements holds:

• • if $l\in [0,n]$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(n)}$ and $\unicode{x3bb} _{\underline {s^{\prime \prime }}(n)}$ have a common endpoint;

• • if $l=n+1$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(n)}\cap \unicode{x3bb} _{\underline {s^{\prime \prime }}(n)}=\emptyset $ .

Then, for the case where $k=n+1$ , we define $\unicode{x3bb} _{\underline {s}(n+1)}$ for any $\underline {s}(n+1)\in \Sigma ^{n+1}_3$ as follows. Note that $\Sigma ^{n}_3\subset \Sigma ^{n+1}_3$ for any $n\geq 0$ . The argument is divided into two cases.

Case A: Suppose $l\in [0,n]$ . It follows from Lemma 5.5 that $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^\infty _3}$ . In particular, $\{\underline {s^{\prime }}(n)\lhd \underline {s^{\prime \prime }}(n)\}_{\Sigma ^{n+1}_3}$ . Set $\underline {s^{\prime }}(n+1)=\underline {s^{\prime }}(n)$ and $\underline {s^{\prime \prime }}(n+1)=\underline {s^{\prime \prime }}(n)$ . Then $\{\underline {s^{\prime }}(n+1)\lhd \underline {s^{\prime \prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ . Define

(5.13)

Case B: Suppose $l=n+1$ . It follows from Lemma 5.5 that there are exactly two elements $\underline {t^{\prime }}(n+1),\underline {t^{\prime \prime }}(n+1)\in \Sigma ^{n+1}_3\setminus \Sigma ^n_3$ such that

$$ \begin{align*}\{\underline{s^{\prime}}(n)\lhd\underline{t^{\prime}}(n+1)\lhd\underline{t^{\prime\prime}}(n+1)\lhd\underline{s^{\prime\prime}}(n)\}_{\Sigma^{n+1}_3}.\end{align*} $$

We first define $\unicode{x3bb} _{\underline {s^{\prime }}(n+1)}$ and $\unicode{x3bb} _{\underline {s^{\prime \prime }}(n+1)}$ as (5.13). Then, we define $\unicode{x3bb} _{\underline {t^{\prime }}(n+1)}$ and $\unicode{x3bb} _{\underline {t^{\prime \prime }}(n+1)}$ as follows. Since $\unicode{x3bb} _{\underline {s^{\prime }}(n+1)}\cap \unicode{x3bb} _{\underline {s^{\prime \prime }}(n+1)}=\emptyset $ , we divide the middle one-third of the shorter component between $\unicode{x3bb} _{\underline {s^{\prime }}(n+1)}$ and $\unicode{x3bb} _{\underline {s^{\prime \prime }}(n+1)}$ into two parts with the same length. One part of it is defined to be $\unicode{x3bb} _{\underline {t^{\prime }}(n+1)}$ and the other part is defined to be $\unicode{x3bb} _{\underline {t^{\prime \prime }}(n+1)}$ such that $\unicode{x3bb} _{\underline {s^{\prime }}(n)},\unicode{x3bb} _{\underline {t}^{\prime }(n+1)},\unicode{x3bb} _{\underline {t}^{\prime \prime }(n+1)},\unicode{x3bb} _{\underline {s^{\prime \prime }}(n)}$ are ordered in the counter-clockwise direction in $\mathbb {S}$ .

Now, let us check that for any $\underline {s^{\prime }}(n+1), \underline {s^{\prime \prime }}(n+1)\in \Sigma ^{n+1}_3$ with $\{\underline {s^{\prime }}(n+1)\lhd \underline {s^{\prime \prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ , one of the following statements holds:

• • if $l\in [0,n+1]$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(n+1)}$ and $\unicode{x3bb} _{\underline {s^{\prime \prime }}(n+1)}$ have a common endpoint;

• • if $l=n+2$ , then $\unicode{x3bb} _{\underline {s^{\prime }}(n+1)}\cap \unicode{x3bb} _{\underline {s^{\prime \prime }}(n+1)}=\emptyset $ .

On one hand, suppose $l\in [0,n+1]$ . According to Lemma 5.5, if $l\in [0,n]$ , it corresponds to $\{\underline {s^{\prime }}(n+1)\lhd \underline {s^{\prime \prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ in Case A; otherwise, if $l=n+1$ , it corresponds to $\{\underline {t^{\prime }}(n+1)\lhd \underline {t^{\prime \prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ in Case B. In both cases, they have a common endpoint by the construction. On the other hand, suppose $l=n+2$ . According to Lemma 5.5, it corresponds to $\{\underline {s^{\prime }}(n+1)\lhd \underline {t^{\prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ in Case A and $\{\underline {t^{\prime \prime }}(n+1)\lhd \underline {s^{\prime \prime }}(n+1)\}_{\Sigma ^{n+1}_3}$ in Case B. In both cases, the intersections of them are empty. This finishes the definition of $\unicode{x3bb} _{\underline {s}}$ for all $\underline {s}\in \bigcup _{n\geq 0}\Sigma ^n_3$ . Note that according to Lemma 5.5, the above construction makes the order ‘ $\lhd $ ’ of $\underline {s}\in \bigcup _{n\geq 0}\Sigma ^n_3$ correspond to the anticlockwise order of $\unicode{x3bb} _{\underline {s}}\in \mathbb {S}$ . See Figure 9 for a sketch.

In addition, for $n\geq 0$ , let

Note that the above sets are mutually disjoint and dense in $\mathbb {S}$ , and hence, induces a partition of $\mathbb {S}$ . In fact, for a set $A\subseteq \mathbb {S}$ , denote by $\mathrm {Length}(A)$ the Euclidean length of A. Note that $\mathrm {Length}(\Lambda _0)=\pi $ and $\mathrm {Length}(\Lambda _1)=\pi /3$ . Moreover, we have

$$ \begin{align*}\mathrm{Length}(\Lambda_{n+1})=2\cdot \mathrm{Length}(\Lambda_{n})/3\end{align*} $$

for any $n\geq 1$ , since the components of $\Lambda _{n+1}$ are twice the number of components of $\Lambda _n$ , while each component of $\Lambda _{n+1}$ is one-third the length of a component of $\Lambda _n$ . Hence, it follows that

$$ \begin{align*} \sum_{n=0}^\infty \mathrm{Length}(\Lambda_{n})=\pi+\pi/3+2\pi/9+4\pi/27+\cdots=2\pi. \end{align*} $$

Part II: the construction of the homeomorphism H. For any $\underline {s}\in \bigcup _{n\geq 0}\Sigma ^n_3$ , define H from $\unicode{x3bb} _{\underline {s}}$ to $\overline {\xi _{\underline {s}}}$ to be a homeomorphism such that by choosing a suitable orientation, H becomes a homeomorphism from $\bigcup _{n\geq 0}\Lambda _{n}$ to $\bigcup _{n\geq 0}\overline {\Gamma _n}$ . It is worth pointing out that we have only proved that H is continuous on one side of every endpoint of $\bigcup _{n\geq 0}\Lambda _n$ . The continuity of H on the other side of these points can be derived by a similar argument as points in $\mathbb {S}\setminus \big(\bigcup _{n\geq 0}\Lambda _{n}\big)$ (see the last paragraph of the proof).

Let $z\in \mathbb {S}\setminus \big(\bigcup _{n\geq 0}\Lambda _{n}\big)$ . We claim that z corresponds to a unique $\underline {s}=s_1s_2\cdots $ with $\underline {s}\in \Sigma _3^\infty \setminus \big(\bigcup _{n\geq 0}\Sigma ^n_3\big)$ such that

(5.14)

On one hand, since the set $\bigcup _{n\geq 0}\Lambda _n$ is dense in $\mathbb {S}$ , it follows from the construction of $\bigcup _{n\geq 0}\Lambda _{n}$ that $z=\lim_{n\rightarrow \infty }\unicode{x3bb} _{\underline {s_n}}$ , where $\underline {s_n}\in \Sigma ^{n}_3$ is unique with