Setting 2.1. Let $\mu $ be a Borel probability measure on the parameter plane ${\mathbb {C}}$ with compact support. Denote the support by $\operatorname {\mathrm {supp}} \mu .$ We define the probability measure $\mathbb {P}_{\mu }$ as the one-sided infinite product of $\mu $ , which is supported on $\Omega _{\mu } = \prod _{n = 1}^{\infty } \operatorname {\mathrm {supp}} \mu $ with the Borel $\sigma $ -algebra $\mathfrak {B}_{\mu }$ .

Proof. By definition, a minimal set $L_{2}$ of $G_{2}$ is forward invariant under $G_{2}$ . Since ${\Gamma _{1} \subset \Gamma _{2}}$ , the set $L_{2}$ is forward invariant also under $G_{1}$ . It follows from Zorn’s lemma that there exists a minimal set $L_{1}$ of $G_{1}$ such that $L_{1} \subset L_{2}$ .

Proof. If $r> {\mathit {r}_{\mathrm {bif}}}(c),$ then $T_{c, r} \equiv 1$ by Theorems 2.16 and 2.21. It follows from Corollary 3.8 that the random Julia set $J_{\omega }$ is totally disconnected almost surely.

Proof. We show that for every $z \in {\widehat {\mathbb {C}}}$ , there exist $c_{1}, c_{2} , \ldots \in {\bar {B}({c}, {r})} $ such that $f_{c_{n}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}} (z) \to \infty $ , which implies that $G_{c, r}$ has no planar minimal sets.

The proof is divided into three cases. Denote by $K(f_{\tilde {c}})$ the autonomous filled Julia set of the superattracting map $f_{\tilde {c}}.$ If $z \not \in K(f_{\tilde {c}})$ , then $ f_{\tilde {c}}^{\circ n} (z) \to \infty $ by definition. If $z \in \partial K(f_{\tilde {c}})$ , then there exists $c_{1} \in {\bar {B}({c}, {r})} $ such that $f_{c_{1}}(z) \not \in K(f_{\tilde {c}})$ and hence $ f_{\tilde {c}}^{\circ n} \circ f_{c_{1}} (z) \to \infty $ . If $z \in \mathrm {int }K(f_{\tilde {c}})$ , then by the no wandering domain theorem, the orbit $ f_{\tilde {c}}^{\circ n} (z) $ converges to the attracting cycle of $0$ . By the assumption, we have $0 \in A_{\omega }.$ Since the non-autonomous basin $A_{\omega }$ is open, there exists an open neighborhood D of $0$ such that for every $z' \in D$ , we have ${{f}_{{\omega }}^{({n} )}} (z') \to \infty $ as $n \to \infty $ . Take $N \in {\mathbb {N}}$ so that $ f_{\tilde {c}}^{\circ N} (z) \in D$ , then ${{f}_{{\omega }}^{({n})}} ( f_{\tilde {c}}^{\circ N} (z)) \to \infty $ as $n \to \infty $ . This completes the proof.

Proof. Fix any $z \in L$ . First, we show $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \subset L.$ By definition, there exists $g_{n} \in G_{c, r}$ such that $g_{n}(0) \to z$ as $n \to \infty .$ Then, for every $g \in G_{c, r}$ , we have $g \circ g_{n}(0) \to g(z)$ as $n \to \infty .$ Since $G_{c, r}$ is a semigroup, we have $g \circ g_{n} \in G_{c, r}$ . Thus, we have $g(z) \in L,$ which implies $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \subset L.$ Next, we show $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \supset L.$ For every open set U which intersects L, take $g_{1} \in G_{c, r}$ such that $g_{1}(0) \in U.$ Since $g_{1}$ is continuous, there exists an open set $U_{0}$ such that $0 \in U_{0}$ and $g_{1}(U_{0}) \subset U.$ Since $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \subset L,$ the autonomous orbit $\{{{{f}}_{{\tilde {c}}}^{\circ {n}}} (z)\}_{n \in {\mathbb {N}}}$ is bounded, and hence $z \in K(f_{\tilde {c}})$ . If $z \in \partial K(f_{\tilde {c}})$ , then there exists $c_{1} \in {\bar {B}({c}, {r})} $ such that $f_{c_{1}}(z) \not \in K(f_{\tilde {c}}),$ but this contradicts $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \subset L.$ Thus, $z \in \mathrm {int } K(f_{\tilde {c}})$ . It follows from the no wandering domain theorem that there exists $N \in {\mathbb {N}}$ such that ${{{f}}_{{\tilde {c}}}^{\circ {N}}} (z) \in U_{0}$ . Thus, $g_{1} \circ {{{f}}_{{\tilde {c}}}^{\circ {N}}} (z) \in U,$ which implies $\overline {\bigcup _{g \in G_{c, r}}\{g(z)\}} \supset L.$ This completes the proof.

Proof. It is obvious that item $(1)$ or $(1')$ is valid. First, we show the equivalence of the four from items $(1)$ to $(4).$

$(1)\Rightarrow (2):$ If $r \leq {\mathit {r}_{\mathrm {bif}}} (c)$ , then ${{f}_{{\omega }}^{({n})}} (0) \not \to \infty $ as $n \to \infty $ for every $\omega \in \Omega _{c, r}$ by Lemma 4.4. Hence, $T_{c, r}(0) = 0$ .

$(2)\Rightarrow (3):$ Take a large $R> 0$ as in Lemma 2.6(1). Suppose that for every $\omega \in \Omega _{c, r}$ , we have $\{{{f}_{{\omega }}^{({n})}}(0) \}_{n=1}^{\infty }$ is bounded. Then $|{{f}_{{\omega }}^{({n})}} (0)| < R$ for every $n\in {\mathbb {N}}$ . By Theorem 3.1, the random Julia set $J_{\omega }$ is connected for every $\omega \in \Omega _{c, r}$ .

$(3)\Rightarrow (4):$ Suppose that the random Julia set $J_{\omega }$ is connected for every $\omega \in \Omega _{c, r}$ . For a large $R> 0$ as in Lemma 2.6(1), it follows from Theorem 3.1 that $|{{f}_{{\sigma ^{k}\omega }}^{({n})}} (0)| \leq R$ for every $\omega \in \Omega _{c, r}$ , $n \in {\mathbb {N}}$ , and $k \in {\mathbb {N}}$ . Thus, the set $L = \overline {\bigcup _{g \in G_{c, r}}\{g(0)\}}$ is bounded in ${\mathbb {C}}$ . By Lemma 4.5, the set L is a planar minimal set.

$(4)\Rightarrow (1):$ This is trivial. See the latter part of Definition 2.17.

The argument above completes the proof of the equivalence from items $(1)$ to $(4).$ Next, we show the equivalence of the four from items $(1')$ to $(4').$

$(1')\Rightarrow (2'):$ If $r> {\mathit {r}_{\mathrm {bif}}} (c)$ , then $G_{c, r}$ has no planar minimal sets by Theorem 2.16. It follows from Theorem 2.21 that $T_{c, r}(z) = 1$ for every $z \in {\widehat {\mathbb {C}}}$ , and hence $T_{c, r}(0) = 1$ .

$(2')\Rightarrow (3'):$ Suppose $T_{c, r}(0) = 1$ . If $r>0$ , then the random Julia set $J_{\omega }$ is totally disconnected ${\mathbb {P}}_{c, r}$ -almost surely by Corollary 3.8. If $r = 0,$ then it trivially follows that the autonomous Julia set is totally disconnected.

$(3')\Rightarrow (4'):$ If item $(3')$ holds, then there exists $\omega \in \Omega _{c, r}$ such that the random Julia set $J_{\omega }$ is disconnected. By Theorem 3.1, there exists $k \in {\mathbb {N}}$ such that ${{f}_{{\sigma ^{k}\omega }}^{({n})}} (0) \to \infty $ as $n \to \infty .$ Since ${{f}_{{\sigma ^{k}\omega }}^{({n})}} \in G_{c, r}$ , then $\overline {\bigcup _{g \in G_{c, r}}\{g(0)\}}$ is unbounded.

$(4')\Rightarrow (1'):$ Let $R> 0$ be large, as in Lemma 2.6(1). If $\overline {\bigcup _{g \in G_{c, r}}\{g(0)\}}$ is not bounded, then there exists $g \in G_{c, r}$ such that $|g(0)|> R.$ By definition of $G_{c, r},$ there exist $N \in {\mathbb {N}}$ and $c_{1}, c_{2}, \ldots , c_{N} \in {\bar {B}({c}, {r})} $ such that $g = {f_{{c_{N}}} \circ \cdots \circ f_{{c_{2}}} \circ f_{{c_{1}}}}.$ Letting $c_{n} = c$ for every $n \geq N+1$ and $\omega = (c_{n})_{n=1}^{\infty }$ , we have ${{f}_{{\omega }}^{({n})}} (0) \to \infty $ as $n \to \infty $ by Lemma 2.6. It follows from Lemma 4.4 that $r> {\mathit {r}_{\mathrm {bif}}} (c)$ .

Proof. It is obvious that the ‘only if’ part is true. Suppose $T_{c, r}(0) = 0$ and we show there does not exist $\omega \in \Omega _{c, r}$ such that ${{f}_{{\omega }}^{({n})}} (0) \to \infty $ as $n \to \infty $ by contradiction. Take a large number $R> 0$ as in Lemma 2.6(1). If there exists $\omega = (c_{n})_{n \in {\mathbb {N}}} \in \Omega _{c, r}$ such that ${{f}_{{\omega }}^{({n})}}(0) \to \infty $ , then there exists $N \in {\mathbb {N}}$ such that $|{f_{{c_{N}}} \circ \cdots \circ f_{{c_{2}}} \circ f_{{c_{1}}}} (0)|> R.$ Since ${\mathbb {C}} \ni c \mapsto f_{c} \in {\mathrm {Poly}}$ is continuous, there exists an open neighborhood $U_{n}$ of $c_{n}$ for every $n =1, 2, \ldots , N$ such that $|{f_{{c^{\prime }_{N}}} \circ \cdots \circ f_{{c^{\prime }_{2}}} \circ f_{{c^{\prime }_{1}}}} (0)|> R$ for every $c^{\prime }_{n} \in U_{n}$ ( $n =1, 2, \ldots , N$ ). Moreover, we have ${\mathbb {P}}_{c, r} ( \mathfrak {U} )> 0$ for $ \mathfrak {U} = \prod _{n=1}^{N} U^{j}_{n} \times \prod _{N+1}^{\infty } {\bar {B}({c}, {r})} .$ However, for every $\omega ' \in \mathfrak {U}$ , it follows from Lemma 2.6 that ${{f}_{{\omega '}}^{({n})}} (0) \to \infty $ as $n \to \infty $ , contradicting the assumption $T_{c, r}(0) = 0$ . This shows that for every $\omega \in \Omega _{c, r}$ , we have $|{{f}_{{\omega }}^{({n})}} (0)| \leq R$ for every $n \in {\mathbb {N}}$ .

Proof. Fix $c' \in {\bar {B}({c}, {r})} \setminus\! {\,\mathcal {M}}$ . Then the autonomous filled Julia set $K(f_{c'})$ of $f_{c'}$ has no interior point. Thus, if $z \in K(f_{c'})$ , then there exists $c_{1} \in {\bar {B}({c}, {r})} $ such that $f_{c_{1}}(z) \notin K(f_{c'})$ . In this case, we have $f_{c'}^{\circ n} \circ f_{c_{1}}(z) \to \infty $ as $n \to \infty .$ Also, if $z \notin K(f_{c'})$ , then by definition, $f_{c'}^{\circ n}(z) \to \infty $ as $n \to \infty .$ This completes the proof.

Proof. From Lemma 4.8, if $r> \operatorname {\mathrm {dist}}(c, \partial\mathcal {M}),$ then the polynomial semigroup $G_{c, r}$ has no planar minimal sets. By Definition 2.17, ${\mathit {r}_{\mathrm {bif}}}(c)$ is the infimum value for $r \geq 0$ such that $G_{c, r}$ has no planar minimal sets. Hence, ${\mathit {r}_{\mathrm {bif}}}(c) \leq \operatorname {\mathrm {dist}} (c, \partial\mathcal {M})$ .

Proof. Fix $c, c' \in {\mathbb {C}}$ . Pick any $r> | c -c' | $ and set $\epsilon = r - | c -c' |> 0$ . By definition, the polynomial semigroup $G_{c, {\mathit {r}_{\mathrm {bif}}}(c)+\epsilon }$ has no planar minimal sets. Since ${\bar {B}({c}, {{\mathit {r}_{\mathrm {bif}}}(c) + \epsilon })} \subset {\bar {B}({c'}, {{\mathit {r}_{\mathrm {bif}}}(c)+r})}$ , it follows from Lemma 2.12 that also $G_{c', {\mathit {r}_{\mathrm {bif}}}(c)+r}$ has no planar minimal sets. This implies that ${\mathit {r}_{\mathrm {bif}}}(c') \leq {\mathit {r}_{\mathrm {bif}}}(c)+r$ , and hence ${\mathit {r}_{\mathrm {bif}}}(c') - {\mathit {r}_{\mathrm {bif}}}(c) \leq | c -c' |.$ Exchanging c and $c'$ with each other, we have also ${\mathit {r}_{\mathrm {bif}}}(c) - {\mathit {r}_{\mathrm {bif}}}(c') \leq | c' -c |,$ which completes the proof.

Proof. This is essentially due to [Reference Brück, Büger and Reitz4]. By Lemma 4.9, we have ${\mathit {r}_{\mathrm {bif}}}(0) \leq 1/4$ . A straight calculation shows that $\overline {D_{1/2}}$ is forward invariant under the polynomial semigroup $G_{0, 1/4}$ , and hence ${\mathit {r}_{\mathrm {bif}}}(0) \geq 1/4.$

Proof. By Lemma 4.9, we have ${\mathit {r}_{\mathrm {bif}}}(c) \leq 1/4 - c$ if $0 \leq c \leq 1/4$ . If $c' = 0$ , then ${\mathit {r}_{\mathrm {bif}}}(c') = 1/4$ from Corollary 4.11. Thus, Theorem 4.10 yields that $|\kern1.5pt{\mathit {r}_{\mathrm {bif}}}(c) - 1/4| \leq c$ for $0 \leq c \leq 1/4$ , which completes the proof.

Proof. Statement (i) follows from the direct calculation. Now, we look at statement (ii). By statement (i), we have $f_{c_{1}}(\overline {D_{\delta }}) \subset \overline {D_{\delta }}$ for every $c_{1} \in {\bar {B}({0}, {r_{0}})}$ . Since ${\bar {B}({0}, {r_{0}})}$ is symmetric under rotation, then the union $\bigcup _{c_{1} \in {\bar {B}({0}, {r_{0}})}} \{f_{c_{1}}(0) \}$ is a round disk centered at the origin. By induction, $\bigcup _{c_{1}, c_{2}, \ldots , c_{n} \in {\bar {B}({0}, {r_{0}})}} \{{f_{{c_{n}}} \circ \cdots \circ f_{{c_{2}}} \circ f_{{c_{1}}}}(0) \}$ is a round disk centered at the origin for every $n \in {\mathbb {N}}$ , and also $\overline {\bigcup _{g \in G_{0, r_{0}}}\{g(0)\}}$ is a closed round disk contained in $\overline {D_{\delta }}$ . If $c_{n}=r_{0}$ for every $n \in {\mathbb {N}}$ , then the autonomous iteration ${{{f}}_{{r_{0}}}^{\circ {n}}}(0)$ converges to the fixed point $\delta $ of $f_{r_{0}}$ . This yields that $\overline {\bigcup _{g \in G_{0, r_{0}}}\{g(0)\}} = \overline {D_{\delta }}$ , and hence $\overline {D_{\delta }}$ is a minimal set of $G_{0, r_{0}}$ .

Proof. Fix any $r> 1/4 + \epsilon - \epsilon ^{2}.$ We show that $G_{-{\epsilon }, r}$ has no planar minimal sets by contradiction, which implies ${\mathit {r}_{\mathrm {bif}}}(-\epsilon ) < r$ . Suppose that there exists a planar minimal set L of $G_{-{\epsilon }, r}$ . Since ${\bar {B}({-\epsilon }, {r})} \supset {\bar {B}({0}, {r- \epsilon })}$ , it follows from Lemma 2.12 that $G_{0, r-\epsilon }$ also has a planar minimal set. By Corollary 4.11, we have $ r- \epsilon \leq 1/4$ . In addition, we assumed that $r - \epsilon> 1/4 - \epsilon ^{2} \geq 0.$ By Lemmas 2.12 and 4.13(ii), we have $L \supset \overline {D_{\delta }}$ for $\delta = (1 - \sqrt {1-4 (r - \epsilon ) })/2$ . Let $z_{0}= i \delta $ and $c_{1} = -r - \epsilon $ , then $z_{0} \in L$ , $c_{1} \in {\bar {B}({-\epsilon }, {r})}$ , and $f_{c_{1}}(z_{0}) = - \delta ^{2} - r - \epsilon = - \delta - 2 \epsilon .$ Now, set $R_{1} = (1 + \sqrt {1 - 4 (r - \epsilon ) })/2> 1/{2}$ , then

$$ \begin{align*} - f_{c_{1}}(z_{0}) -R_{1} &= 2 \epsilon + \frac{1 - \sqrt{1-4 (r - \epsilon) }}{2} - \frac{1 + \sqrt{1-4 (r - \epsilon) }}{2} =2 \epsilon - \sqrt{1-4 (r - \epsilon)}> 0 \end{align*} $$

since $r> 1/4 + \epsilon - \epsilon ^{2}$ . Set $z_{1} = f_{c_{1}}(z_{0})$ , then $z_{1} < - R_{1}$ . Let $c_{n}=r - \epsilon \in {\bar {B}({-\epsilon }, {r})}$ for every $n \geq 2$ . Since $z_{1} < - R_{1}$ , the direct calculation shows $f_{c_{2}}(z_{1})> R_{1}^{2} + r - \epsilon = R_{1}.$ Set $z_{2} = f_{c_{2}}(z_{1})$ and $\alpha = z_{2} - R_{1}$ , then $z_{2} = R_{1} + \alpha $ and $\alpha> 0$ . We can show by induction that ${{{f}}_{{r - \epsilon }}^{\circ {n}}} (z_{2})> R_{1} + \alpha + n \alpha ^{2}$ for every $n \in {\mathbb {N}}$ . Hence, ${{{f}}_{{r - \epsilon }}^{\circ {n}}} (z_{1})$ diverges to $\infty $ as $n \to \infty $ . Thus, we have proved that there exist $z_{0} \in L$ and $\omega = (c_{n})_{n=1}^{\infty } \in \Omega _{-\epsilon , r}$ such that ${{f}_{{\omega }}^{({n})}} (z_{0}) \to \infty $ as ${n \to \infty }.$ This contradicts that L is planar and forward invariant under $G_{-{\epsilon }, r}$ . This completes the proof since r is an arbitrary number that satisfies $r> 1/4 + \epsilon - \epsilon ^{2}.$

Proof. Assume that $f_{c_{2}} \circ f_{c_{1}}(\overline {{D}}_{\delta }) \subset \overline {{D}}_{\delta }$ for every $c_{1}, \ c_{2} \in {\bar {B}({-1}, {r})}$ . For $z = i \delta $ , $c_{1} = -1 -r$ , and $c_{2} = -1 +r$ , we have $f_{c_{1}}(z) = - \delta ^{2} -1-r$ and $f_{c_{2}} \circ f_{c_{1}}(z) = \delta ^{4} +2(1+r)\delta ^{2} + r^{2} + 3r.$ Hence, inequality (1) holds.

Conversely, assume that inequality (1) holds. If $z \in \overline {{D}}_{\delta }$ and $c_{1} \in {\bar {B}({-1}, {r})}$ , then $|f_{c_{1}}(z) +1| \leq |z|^{2} + |c_{1}+1| \leq \delta ^{2} + r$ . In addition, if $|z_{1} + 1| \leq \delta ^{2} + r$ and $c_{2} \in {\bar {B}({-1}, {r})}$ , then

$$ \begin{align*} |f_{c_{2}}(z_{1})| &= |(z_{1} + 1)^{2} -2 z_{1} -2 + 1 + c_{2}| \\ &\leq |z_{1} +1|^{2} + 2|z_{1} + 1| + | 1 + c_{2} | \leq \delta^{4}+2(1+r)\delta^{2} + r^{2} + 3r. \end{align*} $$

Thus, if inequality (1) holds, then $f_{c_{2}} \circ f_{c_{1}}(\overline {{D}}_{\delta }) \subset \overline {{D}}_{\delta }$ for every $c_{1}, \ c_{2} \in {\bar {B}({-1}, {r})}$ .

Proof. The former part $\delta ^{4}+2(1+\rho (\delta ) )\delta ^{2} + \rho (\delta ) ^{2} + 3\rho (\delta ) = \delta $ follows from a straight calculation. We show $\rho (\delta ) \geq 0.$ A simple calculation shows that

$$ \begin{align*} 2 \rho (\delta) = \frac{- ({2 \delta^{2} + 3})^{2} + 4 \delta^{2} + 4\delta +9}{(2 \delta^{2} + 3) + \sqrt{4 \delta^{2} + 4\delta +9}} = \frac{- 4 \delta (\delta^{3} + 2\delta-1)}{(2 \delta^{2} + 3) + \sqrt{4 \delta^{2} + 4\delta +9}}. \end{align*} $$

Define $\tilde {\rho }(\delta ) = \delta ^{3} + 2\delta -1,$ then $\tilde {\rho }(0) = -1 <0$ and $\tilde {\rho }'(\delta ) = 3 \delta ^{2} + 2> 0$ . Thus, ${\rho (\delta ) \geq 0}$ for every $\delta \in [0, \delta _{\mathrm \max }] $ .

Proof. We have that the derivative $\rho '(\delta ) = -2 \delta +(2 \delta + 1)/\sqrt {4 \delta ^{2} + 4 \delta + 9}.$ Thus, $\rho '(\delta ) = 0$ if and only if $16 \delta ^4 + 16 \delta ^3 + 32\delta ^2 - 4 \delta - 1 = 0$ . Let $\delta ^{*} = 0.229\cdots $ be the unique positive root of this quartic equation. Then, $\rho (\delta )$ takes the maximum value at $\delta = \delta ^{*}$ . A numerical computation shows that the maximum value $r_{\mathrm \max }$ is approximately $0.0386\cdots $ .

By Lemmas 4.16 and 4.17, for every $\delta \in [0, \delta _{\mathrm \max }] $ , the set $\overline {{D}}_{\delta } $ gives a forward invariant set under the polynomial semigroup $G_{-1, \rho (\delta )}$ . Thus, we have $\rho (\delta ) \leq r_{\mathrm {bif}}(-1)$ . In particular, $r_{\mathrm \max } = \rho (\delta ^{*}) \leq r_{\mathrm {bif}}(-1)$ . This completes the proof.

Proof. Fix $r> r_{0}$ and apply Lemma 4.4. Set $g = f_{c_{N}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}}$ and let $z^{*}$ denote the parabolic periodic point of g. Then there exists $p\in {\mathbb {N}}$ such that ${{{g}}^{\circ {p}}} (z^{*}) = z^{*}$ , and the multiplier of ${{{g}}^{\circ {p}}}$ at $z^{*}$ is $1.$ Thus, there exists a critical point $z_{0}$ of g such that ${{{g}}^{\circ {np}}}(z_{0}) \to z^{*}$ as $n \to \infty .$ Since $g = f_{c_{N}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}}$ , there exists $0 \leq k \leq N-1$ such that $f_{c_{k}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}} (z_{0}) = 0$ by the chain rule. Here, if $k =0 $ , then we define $f_{c_{k}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}}$ to be the identity map. In any case, we have ${{{g}}^{\circ {np-1}}} \circ {f_{{c_{N}}} \circ \cdots \circ f_{{c_{k+2}}} \circ f_{{c_{k+1}}}} (0) \to z^{*}$ as $n \to \infty .$ Now, let $\epsilon = r - r_{0}> 0$ , then there exists $n_{0} \in {\mathbb {N}}$ such that $n_{0} \geq 2$ and

$$ \begin{align*} |{{{g}}^{\circ {n_{0}p-1}}} \circ {f_{{c_{N}}} \circ \cdots \circ f_{{c_{k+2}}} \circ f_{{c_{k+1}}}} (0) - z^{*} | < \epsilon. \end{align*} $$

Set $w = {{{g}}^{\circ {n_{0}p-1}}} \circ {f_{{c_{N}}} \circ \cdots \circ f_{{c_{k+2}}} \circ f_{{c_{k+1}}}} (0)$ . Since the parabolic periodic point $z^{*}$ belongs to the autonomous Julia set $J(g)$ of g and the Julia set is the boundary of the basin at infinity, perturbation of $c_{N}$ allows us to find a random orbit which escapes to the basin at infinity. Namely, there exists $c' \in {\bar {B}({c_{N}}, {\epsilon })}$ such that

$$ \begin{align*} w' = f_{c'} \circ f_{c_{N-1}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}} \circ {{{g}}^{\circ {n_{0}p-2}}} \circ {f_{{c_{N}}} \circ \cdots \circ f_{{c_{k+2}}} \circ f_{{c_{k+1}}}} (0) \in A(g), \end{align*} $$

where $A(g)$ denotes the basin at infinity for the autonomous iteration of g. Note that ${c' \in {\bar {B}({c}, {r})}} $ since $c_{N} \in {\bar {B}({c}, {r_{0}})}$ and $\epsilon = r - r_{0}$ . By definition, we have ${{{g}}^{\circ {n}}} (w') \to \infty $ as $n \to \infty $ . Thus, Lemma 4.4 gives that ${\mathit {r}_{\mathrm {bif}}}(c) < r.$ This completes the proof since r is an arbitrary number that satisfies $r>r_{0}.$

Proof. Under the assumptions, if $\rho = \rho _{0},$ then $f_{c_{N}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}}(z) - z =0$ has a multiple root. This implies that the map $f_{c_{N}} \circ \cdots \circ f_{c_{2}} \circ f_{c_{1}}$ has a parabolic fixed point with multiplier $1.$ Since $c_{1}, c_{2} \ldots , c_{N} \in {\bar {B}({c}, {r_{0}})}$ with $r_{0} = |\rho _{0}|,$ we have $r_{\mathrm {bif}}(c) \leq r_{0}$ by Theorem 4.19.

Proof. By Theorem 4.6, ${\mathbb {P}}_{c, r}$ -almost every random Julia set is totally disconnected if ${r> {\mathit {r}_{\mathrm {bif}}}(c).}$ In addition, we have ${\mathit {r}_{\mathrm {bif}}}(c + \epsilon ) \leq {\mathit {r}_{\mathrm {bif}}}(c) + |\epsilon |$ for every $\epsilon \in {\mathbb {C}}$ by Theorem 4.10. Thus, the conclusion follows from Examples 4.21 and 4.23.

Proof. By Definition 2.17 and Theorem 4.6, the polynomial semigroup $G_{c, r}$ has the planar attracting minimal set $L = \overline {\bigcup _{g \in G_{c, r}}\{g(0)\}}.$ Define $\delta = \operatorname {\mathrm {dist}} (L, J(G_{c,r})),$ which is strictly positive by Definition 2.13 of attracting minimal sets. For every $\omega \in {\bar {B}({c}, {r})} ^{{\mathbb {N}}}$ , $m \geq 0$ , and $n\geq m$ , we have ${{f}_{{\sigma ^{m} \omega }}^{({n - m})}} (0) \in L$ since ${{f}_{{\sigma ^{m} \omega }}^{({n - m})}} \in G_{c, r}$ . Also, $J_{\sigma ^{m} \omega } \subset J(G_{c,r})$ by Proposition 2.9. Therefore, $\omega $ has post-critical distance $\geq \delta $ with uniform $\delta> 0.$ Since the $\ell ^{\infty }$ -ball ${\bar {B}({c}, {r})} ^{{\mathbb {N}}}$ is connected, the ball ${\bar {B}({c}, {r})} ^{{\mathbb {N}}}$ is contained in a hyperbolic component. Thus, every two $\omega , \omega ' \in {\bar {B}({c}, {r})} ^{{\mathbb {N}}}$ are quasiconformally conjugate on their iterated Julia sets by Theorem 4.28.

Proof. First suppose $r < {\mathit {r}_{\mathrm {bif}}} (c)$ . By Theorem 4.29, there exists a quasiconformal conjugacy $\{\varphi _{n}\}_{n \geq 0}$ between $\tilde {\omega }$ and $\omega $ for every $\omega \in {\bar {B}({c}, {r})} ^{{\mathbb {N}}}$ . By definition, the map $\varphi _{0}$ quasiconformally maps $J_{\tilde {c}}$ onto $J_{\omega }.$ Next suppose $r> {\mathit {r}_{\mathrm {bif}}} (c)$ . Then by Theorem 4.6, the non-autonomous Julia set $J_{\omega }$ is totally disconnected almost surely. Since the autonomous Julia set $J_{\tilde {c}}$ of superattracting parameter $\tilde {c}$ is connected, these $J_{\tilde {c}}$ and $J_{\omega }$ cannot be homeomorphic each other.

Question 5.1. If c does not belong to the main cardioid, then does ${\mathit {r}_{\mathrm {bif}}}(c) < \operatorname {\mathrm {dist}}(c, \partial\mathcal {M})$ hold? If $c \notin [0, 1/4],$ then does it hold?

Question 5.2. What is the exact value of ${\mathit {r}_{\mathrm {bif}}}(\tilde {c})$ for a superattracting parameter $\tilde {c}$ ? How does it relate to other dynamical quantities?

Question 5.3. Let $\tilde {c}_{p}$ be a superattracting parameter with period p for every $p \in {\mathbb {N}}$ . How does ${\mathit {r}_{\mathrm {bif}}}(\tilde {c}_{p})$ decrease as $p \to \infty $ ?

Question 5.4. For every superattracting parameter $\tilde {c}$ , is the bifurcation radius ${\mathit {r}_{\mathrm {bif}}}(\tilde {c})$ an algebraic number?

Question 5.5. Can we generalize Theorem 4.6 when ${\bar {B}({c}, {r})}$ does not contain superattracting parameters?

Question 5.6. How does ${\mathit {r}_{\mathrm {bif}}}(c)$ decrease as $c \to \partial {\mathcal {M}}$ ? For example, within $\epsilon> 0$ , how does ${\mathit {r}_{\mathrm {bif}}}(-3/4 + \epsilon )$ decrease as $\epsilon \to 0$ ?

Question 5.7. How does the typical dimension $d_{c, r}$ depend on c and r?

Question 5.8. Suppose $r> {\mathit {r}_{\mathrm {bif}}}(c).$ Then for ${\mathbb {P}}_{c,r} \times {\mathbb {P}}_{c,r}$ -almost every $(\omega , \omega ')$ , are the Julia sets $J_{\omega }$ and $J_{\omega '}$ quasiconformally equivalent? Note that almost every Julia set is homeomorphic to the (usual) Cantor set.

Question 5.9. If $r = {\mathit {r}_{\mathrm {bif}}}(c),$ then is almost every random Julia set $J_{\omega }$ quasiconformally equivalent to the autonomous Julia set?

Proof. Suppose that there exists $c \in {\mathbb {C}}$ such that $\operatorname {\mathrm {supp}} \mu \subset {\bar {B}({c}, {{\mathit {r}_{\mathrm {bif}}}(c)})}$ . By definition, the polynomial semigroup $G_{c, {\mathit {r}_{\mathrm {bif}}}(c)}$ has a planar minimal set. Thus, Lemma 2.12 gives that $G_{\mu }$ also has a planar minimal set.

Similarly, if there exists $c \in {\mathbb {C}}$ such that $\mathrm {int} (\operatorname {\mathrm {supp}} \mu ) \supset {\bar {B}({c}, {{\mathit {r}_{\mathrm {bif}}}(c)})}$ , then $G_{\mu }$ does not have any planar minimal sets. Thus, $G_{\mu }$ is mean stable, and $T_{\{\infty \}, \mu }$ is identically equal to $1$ by Theorem 2.21. Combining this with Theorem 3.7, we have that the random Julia set $J_{\omega }$ is totally disconnected for ${\mathbb {P}}_{\mu }$ -almost every $\omega \in \Omega _{\mu }$ .