Proof. Let $K\in \mathscr {K}(X)$ be a strict attractor of F. Then there exists an open set $U\subset X$ such that $K\subset U$ and $F^{n}(S)\to K$ in the Hausdorff distance for all $S\in \mathscr {K}(U)$ . According to [Reference SrivastavaSri08, Proposition 2.4.14], since U is an open set of X, then $\mathscr {K}(U)$ is an open set in $\mathscr {K}(X)$ . Therefore, K is an attractor of F.

Proof of Theorem D

Since X is an attractor of F, there exists $\delta>0$ such that

(3) $$ \begin{align} \lim_{n\to\infty}{\mathit{d}_{\mathrm{H}}(F^{n}(S),X)=0} \quad \text{for all } S\in B(X,\delta). \end{align} $$

If X is a singleton, then it is obviously stable. Suppose that X is not a singleton. We will proceed by contradiction. Assume that X is not stable. Hence, there exist $\varepsilon>0$ and sequences $(k_{n})_{n\in \mathbb {N}} \subset \mathbb {N}$ and $(S_{n})_{n\in \mathbb {N}}\subset \mathscr {K}(X)$ such that

(4) $$ \begin{align} \mathit{d}_{\mathrm{H}}(S_{n},X)<\min \bigg\{\frac{1}{n},\frac{\delta}{2} \bigg\} \quad \text{and} \quad \mathit{d}_{\mathrm{H}}(F^{k_{n}}(S_{n}),X)>\varepsilon. \end{align} $$

Observe that $k_{n}\to \infty $ . Otherwise, F might not be continuous. Since F is continuous, for any $n\in \mathbb {N}$ , the set $\mathcal {B}_{n}\subset \mathscr {K}(X)$ defined by

$$ \begin{align*}\mathcal{B}_{n}=\{S\in B(X,\delta/2): \mathit{d}_{\mathrm{H}}(F^{k_{n}}(S),X)>\varepsilon\}\end{align*} $$

is open and non-empty, since $S_{n}\in \mathcal {B}_{n}$ .

Before proving the above claim, let us explain how to use it to conclude the proof of the theorem. Given $n_{1}\in \mathbb {N}$ , since $\mathcal {B}^{1}\ {\stackrel {\scriptscriptstyle \mathrm {def}}{=}}\ \mathcal {B}_{n_{1}}$ is a non-empty open set, by Claim 2.8, there is $n_{2}>n_{1}$ such that $\mathcal {B}^1\cap \mathcal {B}_{n_{2}}\neq \emptyset $ . Applying the claim again to the non-empty open set $\mathcal {B}^{2}\ {\stackrel {\scriptscriptstyle \mathrm {def}}{=}}\ \mathcal {B}^{1}\cap \mathcal {B}_{n_{2}}\subset \mathcal {B}_{n_{2}}$ , we obtain $n_{3}>n_{2}$ such that $\mathcal {B}^{2}\cap \mathcal {B}_{n_{3}}\neq \emptyset $ . Repeating this process inductively, we obtain a sequence $(\mathcal {B}^n)_{n\geq 1}$ of non-empty open sets in $\mathscr {K}(X)$ . Observe that this sequence is nested, that is, $\mathcal {B}^{n+1}\subset \mathcal {B}^n$ for all $n\geq 1$ . Moreover, for any $i\in \mathbb {N}$ , we also have

$$ \begin{align*} \mathit{d}_{\mathrm{H}}(F^{k_{n_{i}}}(S),X)>\varepsilon \quad\text{with}\quad S\in\mathcal{B}^{i}\subset\mathcal{B}_{n_{i}}. \end{align*} $$

Let $\overline {\mathcal {B}^n}$ be the closure (in Hausdorff topology) of $\mathcal {B}^n$ . Since X is compact, we have that $\mathscr {K}(X)$ is also compact (cf. [Reference SrivastavaSri08, Theorem 2.4.17]). Thus, $\overline {\mathcal {B}^{n}}$ is a compact subset of $\mathscr {K}(X)$ for any n. Since the sequence $(\overline {\mathcal B^n})_{n\geq 1}$ is also nested, the intersection $\overline {\mathcal {B}^{\infty }}\ {\stackrel {\scriptscriptstyle \mathrm {def}}{=}}\ \bigcap _{n=1}^{\infty }{\overline {\mathcal {B}^{n}}}$ is not empty. As F is continuous, we have

$$ \begin{align*} \mathit{d}_{\mathrm{H}}(F^{k_{n_{i}}}(S),X)\geq\varepsilon,\quad i\in\mathbb{N},\ S\in\overline{\mathcal{B}^{\infty}}\subset B(X,\delta). \end{align*} $$

This contradicts equation (3). Therefore, X is stable (and then asymptotically stable since it is also an attractor).

Moreover, as the Hutchinson operator is a continuous and monotonic map defined on the hyperspace of compact sets, we can apply the first part of this theorem when X is a strict attractor of an IFS. This completes the proof of Theorem D.

Proof of Claim 2.8

Since $\mathcal {B}$ is a non-empty open set, there are $C\in \mathcal {B}$ and $r>0$ such that $B(C,r)\subset \mathcal {B}$ . Consider $B_{{r}/{4}}(C)=\{x\in X:~\mathit {d}(x,C)<r/4\}$ . The closure $\overline {B_{{r}/{4}}(C)}$ (with respect to the topology induced by $\mathit {d}$ ) is a closed subset of X, and since X is compact, $\overline {B_{{r}/{4}}(C)}$ is itself a compact subset of X.

Observe that $\overline {B_{{r}/{4}}(C)}$ is a subset of the union of balls $B(x,r/4)=\{y\in X:~\mathit {d}(x,y)<r/4\}$ , where x is in $\overline {B_{{r}/{4}}(C)}$ . By compactness, we can choose $x_1,\ldots ,x_p \in \overline {B_{{r}/{4}}(C)}$ such that

(5) $$ \begin{align} \overline{B_{{r}/{4}}(C)}\subset\bigcup_{i=1}^{p}{B(x_i,r/4)}. \end{align} $$

For any $z\in {B(x_1,r/4)}\cup \cdots \cup {B(x_p,r/4)} $ , there is $j\in \{1,\ldots ,p\}$ such that $\mathit {d}(z,x_{j})<r/4$ . Moreover, since $x_{j}\in \overline {B_{{r}/{4}}(C)}$ , we have $\mathit {d}(x_{j},C)\leq r /4$ , which implies that $\mathit {d}(z,C)<r/2$ . Therefore,

(6) $$ \begin{align} \bigcup_{i=1}^{p}{B(x_i,r/4)}\subset B_{{r}/{2}}(C). \end{align} $$

However, from equation (4), we have that $\mathit {d}_{\mathrm {H}}(S_{m},X)\to 0$ as $m\to \infty $ . Then there is $m>n$ such that $S_{m}\cap B(x_i,r/4)\neq \emptyset $ for all $i=1,\ldots ,p$ . Take $S^{\prime }_{m}= S_{m} \cap \overline {B_{{r}/{2}}(C)} \in \mathscr {K}(X)$ . From equation (6), we have:

1. (i) $S^{\prime }_{m}\cap B(x_i,r/4)\neq \emptyset ~\text {for all}~i=1,\ldots ,p$ ;

2. (ii) $S^{\prime }_{m}\subset B_{r}(C)$ .

We can prove two additional statements:

1. (a) $\mathit {d}_{\mathrm {H}}(S^{\prime }_{m},C)<r$ ;

2. (b) $\mathit {d}_{\mathrm {H}}(F^{k_{m}}(S^{\prime }_{m}),X)>\varepsilon $ .

To see statement (a), we use equation (5) to obtain $C\subset \overline {B_{{r}/{4}}(C)}\subset B(x_1,r/4)\cup \cdots \cup B(x_p,r/4)$ . For any $c \in C$ , there exists $j \in {1,\ldots ,p}$ such that $c \in B(x_j,r/4)$ . By item (i), we can find $x' \in B(x_j,r/4) \cap S^{\prime }_{m}$ , and then by the triangle inequality, we get $\mathit {d}(c,x')<r/2$ . Thus, $c \in B_{{r}/{2}}(S^{\prime }_{m})$ , which implies $C \subset B_{{r}/{2}}(S^{\prime }_{m})$ . By combining this with item (ii), we obtain statement (a).

To prove statement (b), we note that since $S_{m}\in \mathcal {B}_{m}$ , we have $\mathit {d}_{\mathrm {H}}(F^{k_{m}}(S_{m}),X)>\varepsilon $ . Since $S^{\prime }_{m}\subset S_{m}$ and F is monotone, we have $\mathit {d}_{\mathrm {H}}(F^{k_{m}}(S^{\prime }_{m}),X)\geq \mathit {d}_{\mathrm {H}}(F^{k_{m}}(S_{m}),X)>\varepsilon $ .

To conclude the proof of the claim, we use statement (a) to obtain

(7) $$ \begin{align} S^{\prime}_{m}\in B(C,r) \subset\mathcal{B}\subset\mathcal{B}_{n}=\{S\in B(X,\delta/2):\mathit{d}_{\mathrm{H}}(F^{k_{n}}(S),X)>\varepsilon\}. \end{align} $$

In particular, $S^{\prime }_m \in B(X,\delta /2)$ and, using statement (b), we conclude that $S^{\prime }_m \in \mathcal {B}_m$ . Therefore, combining this with equation (7), we have $S^{\prime }_m \in \mathcal {B} \cap \mathcal {B}_m$ .

Proof. Since X is a strict attractor of F by Proposition 2.4, X is an attractor of F. Thus, by Theorem D, we have that X is stable. Then, given $\varepsilon>0$ , there is $\delta>0$ such that for any $S\in \mathscr {K}(X)$ with $\mathit {d}_{\mathrm {H}}(S,X)<\delta $ , we have that

$$ \begin{align*} \mathit{d}_{\mathrm{H}}(F^{m}(S),X)<\frac{\varepsilon}{2} \quad \text{for all }\ m\geq0. \end{align*} $$

However, since X is a strict attractor of F, we have $\lim _{n\to \infty }{\mathit {d}_{\mathrm {H}}(F^{n}(S),X)}=0$ for all $S\in \mathscr {K}(X)$ . In particular, given $y\in X$ , we have $ \lim _{n\to \infty }{\mathit {d}_{\mathrm {H}}(F^{n}(\{y\}),X)}=0$ . From this, there exists $n_{y}\in \mathbb {N}$ such that $\mathit {d}_{\mathrm {H}}(F^{n_{y}}(\{y\}),X)<\delta $ . From the continuity of $F^{n_{y}}$ , there is a neighborhood $B_{y}$ of y such that $\mathit {d}_{\mathrm {H}}(F^{n_{y}}(\{x\}),X)<\delta $ for all $x\in B_{y}$ . Thus, we can find a cover of X by neighborhoods of the form $B_{y}$ , that is,

$$ \begin{align*} X\subset \bigcup_{y\in X}{B_{y}}\quad\text{with }\,\mathit{d}_{\mathrm{H}}(F^{n_{y}}(\{x\}),X)<\delta\text{ for all } x\in B_{y}. \end{align*} $$

By the compactness of X, there are $n_{1},n_{2},\ldots ,n_{r}\in \mathbb {N}$ and open sets $B_{1},B_{2},\ldots ,B_{r}$ such that

$$ \begin{align*} X\subset \bigcup_{i=1}^{r}{B_{i}}\quad\text{with }\,\mathit{d}_{\mathrm{H}}(F^{n_{i}}(\{x\}),X)<\delta\text{ for all } x\in B_{i}. \end{align*} $$

From here, by stability of X, we have

$$ \begin{align*} \mathit{d}_{\mathrm{H}}(F^{m}(F^{n_{i}}(\{x\})),X)<\frac{\varepsilon}{2} \quad\text{for all } m\geq 0 \text{ and } x\in B_{i}, \ i=1,\ldots,r. \end{align*} $$

Taking $n_{0}=\max \{n_{1},n_{2},\ldots ,n_{r}\}$ , we have

(8) $$ \begin{align} \text{d}_{H}(F^{n}(\{y\}),X)<\frac{\varepsilon}{2}\quad\text{for all } n\geq n_{0} \text{ and } y\in X. \end{align} $$

From equation (8), we have $X\subset B_{{\varepsilon }/{2}}(F^{n}(\{y\}))$ for all $n\geq n_{0}$ and $y\in X$ . However, note that $B_{{\varepsilon }/{2}}(F^{n}(\{y\}))\subset X$ . So we have $X=B_{{\varepsilon }/{2}}(F^{n}(\{y\}))$ . This immediately implies that $F^{n}(\{y\})$ is $\varepsilon $ -dense on X for all $n\geq n_{0}$ and $y\in X$ .

Proof. Let us prove that $Lf$ is upper semicontinuous, that is, the set $E_a\ {\stackrel {\scriptscriptstyle \mathrm {def}}{=}}\ \{x\in X : Lf(x)<a\}$ is an open set for any $a>0$ . To do this, for a fixed $z\in E_a$ , by definition of $Lf(z)$ , we have $r>0$ such that

$$ \begin{align*}a>L_rf(z)\ {\stackrel{\scriptscriptstyle\mathrm{def}}{=}}\ \mathrm{Lip}(f|_{B(z,r)}) \geq \sup_{x\in B(z,r)} Lf(x) .\end{align*} $$

Thus, $B(z,r) \subset E_a$ and, consequently, $E_a$ is open.

Now we prove the chain rule inequality. Since h is a Lipschitz transformation and thus, a uniform continuous map, given $\rho> 0$ , there exists $r> 0$ such that if $\mathit {d}(x, y) < r$ , then $\mathit {d}(h(x), h(y)) < \rho $ . Hence,

$$ \begin{align*} \frac{\mathit{d}(f(h(x)), f(h(y)))}{\mathit{d}(x, y)} = \frac{\mathit{d}(f(h(x)), f(h(y)))}{\mathit{d}(h(x), h(y))} \cdot \frac{\mathit{d}(h(x), h(y))}{\mathit{d}(x, y)} \leq L_\rho f(h(x)) \cdot L_r h(x). \end{align*} $$

Taking $\rho $ and r tending to $0^+$ , we get the desired chain rule.

Proof. Clearly, equation (9) implies that $\mathscr {F}$ is expanding. For the converse, assume that $\mathscr {F}$ is expanding. For each $x\in X$ , there exists a map $h \in \langle \mathscr {F}\rangle ^+$ such that $Lh(h^{-1}(x))<\rho _x$ with $\rho _x<1$ . By the upper semicontinuity of $Lh$ provided by Lemma 3.3, and since $h^{-1}$ is continuous (as h is a homomorphism), we have $Lh \circ h^{-1} < \rho _x$ in a neighborhood of x. In fact, we can take $\epsilon _x>0$ such that $Lh(z)<\rho _x$ for every $z\in B(h^{-1}(x),{2\epsilon _x})$ . Considering the neighborhood $B_x=h(B(h^{-1}(x),{\epsilon _x}))$ of x, by compactness of X, we get a cover of X by open sets $B_1,\ldots ,B_n$ ; constants $\kappa>1$ , $\epsilon _1,\ldots ,\epsilon _n>0$ , $\epsilon =\min \{\epsilon _1,\ldots ,\epsilon _n\}$ ; and maps $h_1,\ldots ,h_n\in \langle \mathscr {F}\rangle ^+$ such that equation (9) holds with $U_i\ {\stackrel {\scriptscriptstyle \mathrm {def}}{=}}\ B(h^{-1}_i(x_i),{2\epsilon _i})$ , concluding the proof.

Proof. Since $(X,\mathit {d})$ is a locally length space, for each $z\in X$ , there is a neighborhood U of z such that for any point in $x,y\in U$ ,

(10) $$ \begin{align} \mathit{d}(x,y)=\inf\ell(\gamma) \quad \text{where } \gamma:[0,1]\to X \text{ runs over the rectifiable paths from } x \text{ to } y. \end{align} $$

Due to the compactness of X, there exists an open covering $\{U_1, \ldots , U_n\}$ of X such that equation (10) holds for any $x,y \in U_i$ , for $i=1, \ldots , n$ . Let $R> 0$ be a Lebesgue number of this cover. This means that every subset of X with diameter less than $2R$ lies within some member of the cover. Thus, for any $0 < r \leq R$ and any $z \in X$ , the ball $B(z,r)$ is contained in some $U_i$ for $i \in \{1, \ldots , n\}$ , and hence satisfies equation (10) for all $x,y \in B(z,r)$ .

Now, we follow [Reference Durand-Cartagena and JaramilloDCJ10, Lemma 2.2]. Fix $z \in X$ , $0 < r \leq R$ , and choose an arbitrary $\varepsilon> 0$ . Define

$$ \begin{align*} M\ {\stackrel{\scriptscriptstyle\mathrm{def}}{=}}\ \sup_{x \in B(z,r)} Lf(x). \end{align*} $$

For any rectifiable path $\gamma : [0,1] \to X$ with $\gamma (0) = x$ and $\gamma (1) = y$ , both in $B(z,r)$ , and for each $t \in [0,1]$ , there exists $\rho _t> 0$ such that if $p \in B(\gamma (t),\rho _t) \setminus \{\gamma (t)\}$ , then

$$ \begin{align*} \mathit{d}(f(\gamma(t)), f(p)) \leq (M+\varepsilon) \mathit{d}(\gamma(t),p). \end{align*} $$

Since $\gamma $ is continuous, there exists $\delta _t> 0$ such that $(t - \delta _t, t + \delta _t) \subset \gamma ^{-1}(B(\gamma (t),\rho _t))$ . This family of intervals is an open cover of $[0,1]$ and, by compactness, we can choose a finite subcover, say $\{ I_{j} \}_{i=1}^{n}$ , where $I_j=(t_j-\delta _j,t_j+\delta _j)$ . We may assume, refining the subcovering if necessary, that an interval $I_i$ is not contained in $I_j$ for $i \neq j$ . If we relabel the indices of the points $t_i$ in non-decreasing order, we can now choose a point $s_{i,i+1} \in I_{i} \cap I_{i+1} \cap (t_i,t_{i+1})$ for each $i=1,\ldots , n-1$ . Using the auxiliary points that we have just chosen, we deduce that

$$ \begin{align*} \mathit{d}(x,\gamma(t_1)) + \sum_{i=1}^{n-1} ( \mathit{d}(\gamma(t_i),\gamma(s_{i,i+1})) + \mathit{d}(\gamma(s_{i,i+1}),\gamma(t_{i+1})) ) + \mathit{d}(\gamma(t_n),y) \leq \ell(\gamma), \end{align*} $$

and so

$$ \begin{align*} \mathit{d}(f(x), f(y)) \leq (M+\varepsilon) \ell(\gamma). \end{align*} $$

Finally, since this is valid for all $\varepsilon> 0$ and rectifiable path $\gamma $ from x to y, we conclude that $\mathit {d}(f(x), f(y))\leq M ~\mathit {d}(x,y)$ . This implies that $\mathrm {Lip}(f|_{B(z,r)})\leq M$ . The other inequality always holds, and hence we conclude the equality and complete the proof.

Proof. If condition (11) is satisfied, then it is immediate that $\mathscr {F}$ is expanding. For the converse, assume $\mathscr {F}$ is expanding. Referring to Remark 3.10, the expanding property of $\mathscr {F}$ ensures the robustness of equation (9). From the proof of Lemma 3.9, we note that each $U_i$ is an open ball of radius $2\epsilon _i> 0$ for $i = 1,\ldots ,n$ . Moreover, $\epsilon _i$ can be chosen arbitrarily small. Specifically, we can select $\epsilon _i$ such that $0 < \epsilon _i \leq R/2$ , where $R> 0$ is provided by Proposition 3.11. Thus, from Proposition 3.11, it follows that $\mathrm {Lip}(h_i|_{U_i}) = \sup _{x \in U_i} Lh_i(x) < \kappa ^{-1}$ . This implies equation (11), completing the proof.

Proof. Let $\mathscr {F} \in \operatorname {{\mathrm {IFS}}}_k(X)$ be expanding and minimal. By Corollary 3.12, there exist a neighborhood $\mathcal {U}$ of $\mathscr {F}$ , a cover of X by open sets $B_{1},\ldots ,B_{n}$ , open sets $U_1,\ldots , U_n$ , and constants $\kappa>1$ , $\epsilon>0$ such that for any $\mathscr {G}\in U$ , there are maps ${h}_{1},\ldots ,{h}_{n}\in \langle \mathscr {G}\rangle ^+$ satisfying equation (11). Since X is compact, there is a Lebesgue number $\delta>0$ of the open cover $\{B_i:i=1,\ldots ,n\}$ . Take $0<\varepsilon <\min \{\delta ,\epsilon \}$ .

Consider the neighborhood $\mathcal {W}=\mathcal {V}\cap \mathcal {U}$ of $\mathscr {F}$ .

Proof. Denote $y_0=h^{-1}_i(y) \in h_i^{-1}(B_i)$ . Since $\kappa r \leq \varepsilon <\epsilon $ , we have $B(y_0,\kappa r) \subset B_{\epsilon }(h_i^{-1} (B_i)) \subset U_i$ . Now, if $x\in B(y_0,\kappa r)$ , by equation (11),

$$ \begin{align*} \mathit{d}(h_i(x),y)=\mathit{d}(h_i(x),h_i(y_0))<\kappa^{-1}\mathit{d}(x,y_0)<\kappa^{-1} \cdot \kappa r=r. \end{align*} $$

Therefore, we conclude that $h_{i}(B(y_0,\kappa r))\subset B(y,r)$ and complete the proof of the claim.

Finally, we will prove the density of the orbits $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ for all $x\in X$ . According to Claim 3.14 and since $\mathscr {G} \in \mathcal {W}$ , given $x\in X$ , we have that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is $\varepsilon $ -dense. Now, we will show that actually $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is $\kappa ^{-1}\varepsilon $ -dense in X. To do this, take any $y\in X$ and consider the open ball $B(y,\kappa ^{-1}\varepsilon )$ . Since $\kappa ^{-1}\varepsilon <\delta $ , then there is $i\in \{1,2,\ldots ,n\}$ such that $B(y,\kappa ^{-1}\varepsilon )\subset B_{i}$ . Then, by Claim 3.15, we have

$$ \begin{align*} h_{i}(B(h_i^{-1}(y),\kappa{\kappa}^{-1}\varepsilon))\subset B(y,\kappa^{-1}\varepsilon). \end{align*} $$

Since $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is $\varepsilon $ -dense, there is $h\in \langle \mathscr {G}\rangle ^+$ such that $h(x)\in B(h_i^{-1}(y),\varepsilon )=B(h_i^{-1}(y), \kappa {\kappa }^{-1}\varepsilon )$ . Therefore, $(h_{i}\circ h)(x)\in B(y,\kappa ^{-1}\varepsilon )$ . Consequently, the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is $\kappa ^{-1}\varepsilon $ -dense.

Repeating this argument, we can show that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is $\kappa ^{-p}\varepsilon $ -dense for all $p\in \mathbb {N}$ . Since $\kappa ^{-p}\varepsilon \to 0$ as $p\to \infty $ , we find that any orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)$ is dense. That is, $\mathscr {G}$ is minimal.

Proof. We first note that condition (2) in Definition 3.18 is robust in $(\operatorname {{\mathrm {IFS}}}_k(X),\skew5\bar{\mathit{d}}_0)$ , as $\mathit {d}_0$ controls the $C^0$ distance and the locally Lipschitz constant. Next, we will prove that the globalization property $X=T_1(B)\cup \cdots \cup T_s(B)$ is also robust. The proof of the robustness of any other cover in Definitions 3.16 and 3.18 is analogous.

Since $(X,\mathit {d})$ is compact, by Lebesgue’s number lemma, there exists a Lebesgue number $\delta>0$ for the open cover $\{T_j(B): j=1,\ldots ,s\}$ . We cover X by finitely many open balls $B(x_i,\delta /2)$ , $i=1,\ldots ,n$ , and for each i, there exists $j_i$ such that $B(x_i,\delta )\subset T_{j_i}(B)$ . If the boundary $\partial B(x_i,\delta /2)$ is empty, then $B(x_i,\delta /2) \subset \tilde {T}(B)$ for any sufficiently small perturbation $\tilde {T}\in \operatorname {{\mathrm {Diff}^0}}(X)$ of $T_{j_i}$ . We can assume that $\partial B(x_i,\delta /2)$ is not empty for any i, and let

$$ \begin{align*}\eta=\min_{1\leq i\leq n} \mathit{d}(\partial B(x_i,\delta/2),\partial T_{j_i}(B))>0.\end{align*} $$

Moreover, $\eta =\eta (\mathscr {F})$ varies continuously with respect to $\mathscr {F}$ in $\operatorname {{\mathrm {IFS}}}_k(X)$ . Thus, we can find a neighborhood $\mathcal {U}$ of $\mathscr {F}$ in $\operatorname {{\mathrm {IFS}}}_k(X)$ such that $\eta (\mathscr {G})>0$ for all $\mathscr {G}\in \mathcal {U}$ . In particular, $B(x_i,\delta /2) \subset \tilde {T}_{j_i}(B)$ for each $i=1,\ldots ,n$ , where $\tilde {T}_{j_i}$ is the perturbation of $T_{j_i}$ formed by the same finite sequence of generators, assuming that $\mathscr {G}$ is minimally ordered with respect to $\mathscr {F}$ . Since $\{B(x_i,\delta /2): i=1,\ldots ,n\}$ covers X, we have ${X=\tilde {T}_1(B)\cup \cdots \cup \tilde {T}_s(B)}$ , completing the proof.

Proof. Consider $\mathscr {F}\in \operatorname {{\mathrm {IFS}}}_k(X)$ and let $\langle \mathscr {F}\rangle ^+$ denote the semigroup generated by $\mathscr {F}$ . By hypothesis, there exists a globalized blending region B for $\mathscr {F}$ . In fact, Proposition 3.19 implies that B is also a globalized blending region for any $\mathscr {G}$ belonging to a small neighborhood $\mathcal {U}$ of $\mathscr {F}$ in $\operatorname {{\mathrm {IFS}}}_k(X)$ . From Definition 3.18, there are $h_{1},h_{2},\ldots ,h_{m}\in \langle \mathscr {G}\rangle ^+$ and an open set $D\subset X$ such that $\overline {B}\subset D$ and conditions (1) and (2) are satisfied.

Proof. To see this, we will use Hutchinson’s theorem [Reference HutchinsonHut81], which states that the Hutchinson operator $F_{\mathscr {H}}:\mathscr {K}(\overline {D})\to \mathscr {K}(\overline {D})$ associated with the family of contractions $\mathscr {H}=\{h_{1},\ldots ,h_{m}\}$ of $\overline {D}$ is also a contraction. As a consequence, there exists a unique compact set $A\subset \overline {D}$ such that

$$ \begin{align*} A=F_{\mathscr{H}}(A)=\bigcap_{n\in\mathbb{N}} F_{\mathscr{H}}^n(\overline{D}) \quad \text{and} \quad A \subset \overline{\mathcal{O}_{\mathscr{H}}^+(x)} \text{ for all } x\in\overline{D}. \end{align*} $$

Since $B\subset \overline {D}$ and $B\subset F_{\mathscr {H}}(B)$ , it follows that $B\subset A$ . Consequently, we have $B \subset \overline {\mathcal {O}_{\mathscr {H}}^+(x)}$ for all $x\in \overline {D}$ . Finally, as $\langle \mathscr {H}\rangle ^+$ is a subsemigroup of $\langle \mathscr {G}\rangle ^+$ , we can conclude that for every $x\in B$ , it holds that $B\subset \overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(x)}$ , as required.

To show that $\mathscr {G}$ is minimal, that is, $X=\overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(y)}$ for every $y\in X$ , we note that since B is globalized for $\mathscr {G}$ , there exist $T_1,\ldots ,T_s, S_1,\ldots ,S_t \in \langle \mathscr {G}\rangle ^+$ such that

$$ \begin{align*} X=T_1(B)\cup \cdots\cup T_s(B) = S_1^{-1}(B)\cup\cdots \cup S_t^{-1}(B). \end{align*} $$

Given $y\in X$ and any open set U of X, we can find $i\in \{1,\ldots ,s\}$ and $j\in \{1,\ldots ,t\}$ such that $y \in S_{j}^{-1}(B)$ and $U\cap T_i(B)\neq \emptyset $ . Therefore, we have $S_j(y)\in B$ and $T_i^{-1}(U)\cap B$ is an open set in B.

Using Claim 3.21, we obtain $B\subset \overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(S_{j}(y))}$ . In particular, we have $T_i^{-1}(U)\cap B\subset \overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(S_{j}(y))}$ . Hence, there is $L\in \langle \mathscr {G}\rangle ^+$ such that $L\circ S_j(y) \in T_i^{-1}(U)\cap B$ , and so $T_i \circ L \circ S_j(y) \in U$ . This proves that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(y)$ is dense in X, and therefore $\mathscr {G}$ is minimal.

Proof. We begin by presenting the following lemma, which will be used to prove the theorem.

Lemma 3.23. Let B be an open set that satisfies equation (12). Then, for every $x\in B$ , there is a sequence $\{i_n\}_{n\geq 1}$ , $i_n \in \mathbb {N}$ such that

$$ \begin{align*} x=\lim_{n\to\infty} h_{i_1}\circ\cdots\circ h_{i_n}(y) \quad \text{for all } y\in B. \end{align*} $$

Proof. We will recursively define the sequence $\{i_n\}_{n\geq 1}$ . Once we find an integer $i_n>0$ such that

$$ \begin{align*} x\in h_{i_1}\circ\cdots \circ h_{i_n}(B), \end{align*} $$

we deduce from equation (12) that

$$ \begin{align*} x\in h_{i_1}\circ\cdots\circ h_{i_n}(B)\subset \bigcup_{i=1}^{\infty} h_{i_1}\circ\cdots \circ h_{i_n}\circ h_i(B). \end{align*} $$

Thus, we can find $i_{n+1}$ such that $x\in h_{i_1}\circ \cdots \circ h_{i_n}\circ h_{i_{n+1}}(B)$ . Hence, we have built the sequence $\{i_n\}_{n\geq 1}$ such that

$$ \begin{align*} x\in \bigcap_{n\geq 1} h_{i_1}\circ \cdots \circ h_{i_n}(B). \end{align*} $$

Furthermore, since $x\in h_{i_1}\circ \cdots \circ h_{i_n}(B)$ for all $n\in \mathbb {N}$ , we deduce that for any $y\in B$ , it holds that

$$ \begin{align*} \mathit{d}(h_{i_1} \circ \cdots \circ h_{i_n}(y), x) \leq \mathrm{diam}(h_{i_1}\circ \cdots \circ h_{i_n}(B)). \end{align*} $$

Note that, since $x\in B$ and $x\in h_{i_1}\circ \cdots \circ h_{i_n}(B)$ , then the sequence $\{i_n\}_{n\geq 1}$ satisfies that $h_{i_1}\circ \cdots \circ h_{i_n}(B)\cap B\not =\emptyset $ for all $n\in \mathbb {N}$ . Thus, by hypothesis, $\lim _{n\to \infty }\mathrm {diam}(h_{i_1}\circ \cdots \circ h_{i_n} (B))=0$ . This shows that $\lim _{n\to \infty }{\mathit {d}(h_{i_1}\circ \cdots \circ h_{i_n}(y), x)}=0$ proving the lemma.

Now, let us proceed with the proof of the theorem. Given a point $x\in X$ and an open set U of X, since $X=\overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {F}}}}(B)} = \mathcal {O}_{\mathscr {F}}^-(B)$ , we have there are $T,S\in \langle \mathscr {F} \rangle ^+$ such that $S(x) \in B$ and $B\cap T^{-1}(U)$ is open not empty. By Lemma 3.23, any point in B can be approximated by some iterates of the form $h_{i_1}\circ\cdots \circ h_{i_n}(S(x))$ . In particular, there is $h\in \langle \mathscr {F}\rangle ^+$ such that $h\circ S(x)\in T^{-1}(U)\cap B$ . Thus, $T\circ h \circ S(x)\in U$ . This proves that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {F}}}}(x)$ is dense in X and therefore, $\mathscr {F}$ is minimal.

Proof. First, notice that since B is a globalized open set for $\mathscr {F}$ , $B\subset \overline {\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {F}}}}(B)} = \mathcal {O}_{\mathscr {F}}^-(B)$ . Hence, to apply Theorem 3.22, we will show that for any $\{i_n\}_{n\geq 1}$ ,

(14) $$ \begin{align} \mathrm{diam}(h_{i_1}\circ\cdots \circ h_{i_n}(B))\to 0 \quad \text{as } n\to\infty. \end{align} $$

To prove this convergence, observe that any composition $h_{i_1}\circ \cdots \circ h_{i_n}$ sends D to D with contraction rate $\beta ^n$ . Hence, $\mathrm {diam}(h_{i_1}\circ \cdots\circ h_{i_n}(B)) \leq \beta ^n \mathrm {diam}(B)$ and consequently equation (14) holds. Therefore, by Theorem 3.22, $\mathscr {F}$ is minimal. To conclude the proof, we will show that $\mathscr {F}$ is expanding. Thus, by Theorem 3.13, we have that $\mathscr {F}$ is robustly minimal.

Given any $x\in X$ , for the globalized property in Definition 3.16, we have $T^{-1}_j(x)\in B$ for some $j\in \{1,\ldots ,m\}$ . Then, by the cover in equation (13), we can iterate $T_j^{-1}(x)$ by some $h_i^{-1}$ and remain in B. That is, $h^{-1}_i\circ T_j^{-1}(x)\in B$ . Since $Lh_i|_B \leq \beta $ , by Lemma 3.3, we have that

$$ \begin{align*} L(T_j\circ h_{i})((h_{i}^{-1}\circ T_j^{-1})(x)) \leq Lh_{i}(h_{i}^{-1}\circ T^{-1}_j(x)) \cdot LT_{j}(T^{-1}_j(x)) \leq \beta LT_{j}(T^{-1}_j(x)). \end{align*} $$

Since $\beta <1$ , repeating the argument the number of times necessary, we found $g\in \langle \mathscr {F}\rangle ^+$ such that $Lg(g^{-1}(x))<1$ and thus $\mathscr {F}$ is expanding.

Proof. Since $f_2$ is a circle diffeomorphism,

$$ \begin{align*} 1=\mathrm{Leb}(\mathbb{S}^1)=\mathrm{Leb}((f_2)^{-1}(\mathbb{S}^1))=\int |(f_2^{-1})'(z)| \, d\mathrm{Leb}(z), \end{align*} $$

where $\mathrm {Leb}$ is the normalized Lebesgue measure on $\mathbb {S}^1$ . Moreover, since $f_2$ is not a rotation, from the above estimate, there exists a point $z\in \mathbb {S}^1$ such that $|(f_2^{-1})'(z)|>1$ . From here, we have an open interval I containing z such that $|(f_2^{-1})'|_I|>1$ . However, since $f_1$ is an irrational rotation, $f_1^{-1}$ is minimal and thus, for every $x\in \mathbb {S}^1$ , there exists $n\in \mathbb {N}$ such that $f_1^{-n}(x)\in I$ . Then,

$$ \begin{align*}|(g^{-1})'(x)|=|(f_2^{-1})'(f_1^{-n}(x))|\cdot |(f_1^{-n})'(x)|=|(f_2^{-1})'(f_1^{-n}(x))|>1,\end{align*} $$

where $g=f_1^{n}\circ f_2 \in \langle \mathscr {F}\rangle ^+$ . This implies that $\mathscr {F}$ is expanding since it is also minimal (since $f_1$ is an irrational rotation), and we conclude from Theorem 3.13 that $\mathscr {F}$ is also robustly minimal.

Proof of Theorem 4.7

Let $\mathscr {F} \in \operatorname {{\mathrm {IFS}}}_k(X)$ be expanding and transitive. Let $\mathcal {U}$ and $\varepsilon>0$ be the neighborhood of $\mathscr {F}$ and any positive constant sufficiently small provided in the proof of Theorem 3.13 from the expanding property, respectively. The main modification that we need to do is the following. Compare with Claim 3.14.

Fix $\mathscr {G}\in \mathcal {W}=\mathcal {U}\cap \mathcal {V}$ and an open ball V of radius $\varepsilon $ . We will prove the density of the $\mathscr {G}$ -orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ . To do this, according to Claim 4.8, we have first that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is $\varepsilon $ -dense. Now, using the notation introduced in Theorem 3.13, we will show that actually $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is $\kappa ^{-1}\varepsilon $ -dense in X. To do this, take any $y\in X$ and consider the open ball $B(y,\kappa ^{-1}\varepsilon )$ . Since $\kappa ^{-1}\varepsilon <\delta $ , then there is $i\in \{1,2,\ldots ,n\}$ such that $B(y,\kappa ^{-1}\varepsilon )\subset B_{i}$ . Then, by Claim 3.15, we have

$$ \begin{align*} h_{i}(B(h_i^{-1}(y),\kappa{\kappa}^{-1}\varepsilon))\subset B(y,\kappa^{-1}\varepsilon). \end{align*} $$

Since $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is $\varepsilon $ -dense, there is $h\in \langle \mathscr {G}\rangle ^+$ and $x\in V$ such that

$$ \begin{align*}h(x)\in B(h_i^{-1}(y),\varepsilon)=B(h_i^{-1}(y),\kappa{\kappa}^{-1}\varepsilon).\end{align*} $$

Therefore, $(h_{i}\circ h)(x)\in B(y,\kappa ^{-1}\varepsilon )$ . Consequently, the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is $\kappa ^{-1}\varepsilon $ -dense. Repeating this argument, we can show that the orbit $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is $\kappa ^{-p}\varepsilon $ -dense for all $p\in \mathbb {N}$ and consequently, $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(V)$ is dense as we wanted to show.

Proof. Let B be the blending region of $\mathscr {F}$ . Since B is an open set, we have $\delta>0$ and $x \in B$ such that $B(x,\delta )\subset B$ . According to Remark 4.3, $\mathscr {F}$ is forward and backward transitive. Since it is also forward and backward expanding, we can apply Theorem 4.7 for $\mathscr {F}$ and $\mathscr {F}^{-1}$ . Then, for a fixed $0<\varepsilon <\delta $ small enough, we have a neighborhood $\mathcal {V}$ of $\mathscr {F}$ in $\operatorname {{\mathrm {IFS}}}_k(X)$ such that for any $\mathscr {G}\in \mathcal {V}$ , the forward and backward $\mathscr {G}$ -orbits $\operatorname {\mathrm {\mathcal {O}^+_{\mathscr {G}}}}(W)$ and $\mathcal {O}_{\mathscr {G}}^-(W)$ of $W=B(x,\varepsilon )$ are both dense on X. Then, given any pair of open sets U and V, there exists $h,g\in \langle \mathscr {G}\rangle ^+$ such that $h^{-1}(U)\cap W \not =\emptyset $ and $g(V)\cap W \not =\emptyset $ . Since $W\subset B$ and B is a blending region, according to Lemma 3.23, we find $H \in \langle \mathscr {G}\rangle ^+$ such that $H(g(V)\cap W) \cap (h^{-1}(U)\cap W)\not =\emptyset $ . In particular, we get that $f(V) \cap U \not =\emptyset $ , where $f=h\circ H\circ g \in \langle \mathscr {G}\rangle ^+$ . This implies that $\mathscr {G}$ is transitive, concluding the proof.