2 Convenient notation

In what follows, we always assume that $f:X\to X$ is a PCIM and that the topology on X is the one induced by the Euclidean metric. Then, there exist $\lambda \in (0,1)$ and a collection of $N\geq 1$ open nonempty disjoint subintervals $X_1,X_2,\ldots ,X_N$ such that ${X=\bigcup _{i=1}^N\overline {X_i}}$ and

(2.1) $$ \begin{align} |f(x)-f(y)|\leq\lambda\,|x-y|\qquad\forall\,x,y\in X_i\,,\;\;\forall\,i\in\{1,2,\ldots,N\}. \end{align} $$

The real number $\lambda $ is called contraction rate of f, and the elements of the collection $\{X_i\}_{i=1}^N$ are called contraction pieces. We will consider them to be sorted. In particular, let $c_0,c_N$ denote the extreme points of X and $\Delta =\{c_1<c_2<\dots <c_{N-1}\}$ the set of the boundaries of the contraction pieces of f; that is,

$$\begin{align*}X_1=[c_0,c_1)\,,\quad X_2=(c_1,c_2)\,,\quad \dots\;,\quad X_N=(c_{N-1},c_N]. \end{align*}$$

For notational convenience, we assume that $X_1$ and $X_N$ are half-closed, but one may also consider the case where one or both pieces are open by adding $c_0$ and/or $c_N$ to $\Delta $ . In other words, $\Delta $ must contain all the discontinuity points of the map.

As we said before, the attractor $\Lambda $ of f is nonempty, compact, pseudo-invariant, and $\widetilde X$ -minimal provided that f is piecewise monotonic and $D\subset \widetilde X$ . The formal definition of the attractor $\Lambda $ is given by

$$\begin{align*}\Lambda:=\bigcap_{n\geq1}\Lambda_n\,, \;\text{ where } \,\Lambda_1=\overline{f(X\backslash\Delta)}\;\, \text{ and } \;\Lambda_{n+1}=\overline{f(\Lambda_n\backslash\Delta)}\; \text{ for all } n\geq 1. \end{align*}$$

We are interested in computing the topological entropy of $f|(\Lambda \cap \widetilde X)$ and the Hausdorff dimension of $\Lambda $ . For the latter, it will be necessary to recall some concepts. If $\delta \geq 0$ and $E\subset X$ , a $\delta $ -cover of E is a countable collection of subsets of X that covers E and the diameter of each of which is smaller than or equal to $\delta $ . Thus, for any $E\subset X$ , and every $s\geq 0$ and $\delta>0$ , we consider the numbers

$$\begin{align*}\mathcal H_\delta^s(E):=\inf\left\{\sum_{C\in\mathcal C}\big(\!\operatorname{\mathrm{diam}}(C)\big)^s\ :\ \mathcal C \text{ is a } \delta\text{-cover of } E\right\} \end{align*}$$

and

$$\begin{align*}\mathcal H^s(E):=\lim_{\delta\to0}\mathcal H_\delta^s(E). \end{align*}$$

It is known that $\mathcal H^s_\delta $ and $\mathcal H^s$ define an outer measure and a measure on X, respectively. Furthermore, $\mathcal H^s$ is called s-dimensional Hausdorff measure on X. The Hausdorff dimension of a set $E\subset X$ , which we denote by $\dim _{\mathcal H}(E)$ , is the critical value $s\geq 0$ where $\mathcal H^s(E)$ jumps from infinity to zero. This number satisfies, for example, that $\dim _{\mathcal H}(E)=0$ for every $E\subset X$ such that $\#E\leq \aleph _0$ . Moreover, if $s_0:=\dim _{\mathcal H}(E)\geq 1$ is a positive integer, then $\mathcal H^{s_0}$ coincides with the Lebesgue measure on $\mathbb R^{s_0}$ . If $s_0=0$ , then it is known that E is totally disconnected (the reciprocal is not true!). In practical terms, the Hausdorff dimension provides a general notion of the size of a set in a metric space.

Additionally, we can consider another fractal dimension that allows us to describe the “size” of sets. Given $E\subset X$ and $\epsilon>0$ , let $\ell (E,\epsilon )$ be the smallest number of intervals with diameter at most $\epsilon $ covering E. Then, we can define the box dimension of E to be

$$\begin{align*}\dim_B(E):=\liminf_{\epsilon\to0^+}\frac{\log \ell(E,\epsilon)}{\log(1/\epsilon)}. \end{align*}$$

Here, we use the lower limit to avoid problems with the convergence. Strictly speaking, this is usually called the lower box dimension and the box dimension is usually said to exist when the limit $\lim _{\epsilon \to 0^+}$ exists. Next, we give the existing comparison between the Hausdorff and box dimensions.

We recall that the previous result is valid in a general metric space, after adapting the definitions accordingly.

3 Complexity and atoms

Let $\mathbb N$ be the set of natural numbers starting at 0. We say that the sequence $\theta =(\theta _t)_{t\geq 0}\in \{1,\ldots ,N\}^{\mathbb N}$ is the itinerary of a point $x\in \widetilde X$ if, for every $t\in \mathbb N$ and $i\in \{1,\ldots ,N\}$ , we have that $\theta _t=i$ if and only if $f^t(x)\in X_i$ . Also, the complexity function of a sequence $\theta =(\theta _t)_{t\geq 0}$ is the function $p_\theta (n):=\#L_n(\theta )$ defined for every $n\geq 1$ , where

$$\begin{align*}L_n(\theta):=\big\{\theta_t\theta_{t+1}\ldots\theta_{t+n-1}\in\{1,\ldots,N\}^n\ :\ t\geq0\big\}. \end{align*}$$

Thus, $p_\theta (n)$ gives the number of different words of length n contained in $\theta $ . The complexity function of any sequence is a non-decreasing function of n. Also, if there exists $n_0\geq 1$ such that $p_\theta (n_0+1)=p_\theta (n_0)$ , it can be shown that $p_\theta (n)=p_\theta (n_0)$ for all $n\geq n_0$ . Theorem 1.2 is the result of classifying the orbits of points in $\widetilde X$ according to the complexity associated with their itineraries.

Note that it is possible to associate the concept of complexity with each point of $\widetilde X$ via its itinerary. It could happen that the complexity associated with orbits of a PCIM grows exponentially. However, when a PCIM satisfies the separation property, the complexity growth is at most affine for each itinerary of the system and n sufficiently large. Specifically, we have the following result.

Theorem 3.1 [Reference Catsigeras, Guiraud and Meyroneinc2]

Suppose that f satisfies the separation property. Let $x\in \widetilde X$ and $\theta \in \{1,\ldots ,N\}^{\mathbb N}$ be its itinerary, then there exist $m_0\geq 1$ , $\alpha \in \{0,\ldots ,N-1\}$ , and $\beta \in \big \{1,\ldots ,1+m_0(N-1-\alpha )\big \}$ such that

(3.1) $$ \begin{align} p_\theta(n)=\alpha n+\beta\qquad\forall\,n\geq m_0. \end{align} $$

Under the assumptions of Theorem 1.2, it can be proved that, for every $x\in \widetilde X$ , $\omega (x)$ is a periodic orbit if and only if the complexity associated with x is eventually constant; that is, $\alpha =0$ (see Theorem 2.2 in [Reference Calderón, Catsigeras and Guiraud1]). Furthermore, $\omega (x)$ is an $\widetilde X$ -minimal Cantor set if and only if the complexity associated with x is eventually affine with $\alpha \neq 0$ (see Theorem 2.3 in [Reference Calderón, Catsigeras and Guiraud1]). One of the important tools that allowed the establishment of these results is the so-called atom.

Let $\mathcal P(X)$ be the power set of X. For every $i\in \{1,\ldots ,N\}$ , let $F_i:\mathcal P(X)\to \mathcal P(X)$ be defined by $F_i(A)=\overline {f(A\cap X_i)}$ for $A\in \mathcal P(X)$ . Let $n\geq 1$ and $(i_1,\ldots ,i_n)\in \{1,\ldots ,N\}^n$ . We say that

$$\begin{align*}A_{i_1i_2\ldots i_n}:= F_{i_n}\circ F_{i_{n-1}}\circ\cdots\circ F_{i_1}(X) \end{align*}$$

is an atom of generation n if it is nonempty. We denote $\mathcal A_n$ the set of all atoms of generation n. Every atom of generation $n\geq 2$ is contained in an atom of previous generation. Precisely, for all $n\geq 2$ and $(i_1,\ldots ,i_n)\in \{1,\ldots ,N\}^n$ , we have that

$$\begin{align*}A_{i_1\ldots i_n}\subset A_{i_2\ldots i_n}\subset\cdots\subset A_{i_n}. \end{align*}$$

Also, each atom of any generation is a compact interval contained in X. Moreover, note that the attractor of f can be defined in terms of atoms as

(3.2) $$ \begin{align} \Lambda=\bigcap_{n\geq1}\Lambda_n\,,\quad \text{ where }\;\, \Lambda_n=\bigcup_{A\in\mathcal A_n}A\quad\forall\,n\geq1. \end{align} $$

We are interested in relating the code associated with an atom to the itinerary of points in X. The following result is relevant for this purpose.

Next, we list some basic properties of atoms. The proof of each property is straightforward and is left as an exercise to the reader. Recall that $\lambda \in (0,1)$ is the contraction rate of f.

Note that the existence of the sequence of atoms $\{B_k\}_{k \geq 1}$ in Lemma 3.3 is guaranteed by (3.2). Besides, the uniqueness in item 4 of the same lemma follows directly from Lemma 3.2. Finally, the relationship between atoms, orbits, and itineraries is established in the following result.

4 Proof of Theorem 1.3

We say that C is a basic piece of $\Lambda $ if C is an $\widetilde X$ -minimal component of the attractor of f; that is, $C\in \{\mathcal O_1,\ldots ,\mathcal O_{N_1},K_1,\ldots ,K_{N_2}\}$ (see Theorem 1.2). Also, if C is a basic piece of $\Lambda $ , we denote by $\mathcal A_n(C)$ the set of all atoms of generation n that intersect C. Because of (3.2), for every $n\geq 1$ , every basic piece is contained in the union of atoms of generation n.

Lemma 4.1 Suppose that f satisfies the separation property and $D\subset \widetilde X$ . If K is a non-periodic basic piece of $\Lambda $ and $\theta $ the itinerary of a point $x\in K\cap \widetilde X$ , then

$$\begin{align*}\mathcal A_n(K)= \{A_{i_1\ldots i_n}\ :\ i_1\ldots i_n\in L_n(\theta)\}\qquad \forall\,n\geq1. \end{align*}$$

In particular, $\#\mathcal A_n(K) = p_\theta (n)$ .

Proof Let $n\geq 1$ . Consider $A\in \mathcal A_n(K)$ and let $y\in A$ . By the separation property, there exists $\epsilon>0$ such that $\epsilon <\min \!\big \{\!\operatorname {\mathrm {dist}}(A_1,A_2)\ :\ A_1,A_2\in \mathcal A_n\,\text { and }\,A_1\neq A_2\big \}$ , where $\operatorname {\mathrm {dist}}$ denotes the Euclidean metric on X. Next, by the $\widetilde X$ -minimality of K, there exist $m\geq n$ and $x\in K\cap \widetilde X$ such that

$$\begin{align*}|y-f^m(x)|<\min\!\big\{\!\operatorname{\mathrm{diam}}(A)/2,\epsilon\big\}. \end{align*}$$

Thus, we deduce that $f^m(x)\in A$ . If $\theta $ is the itinerary of x, from Lemma 3.4 and item (1) of Lemma 3.2, we deduce that

$$\begin{align*}f^{m}(x)=f^{(m-n)+n}(x)\in A_{\theta_{m-n}\ldots\theta_{m-1}}=A. \end{align*}$$

Then, we conclude that $y\in A_{\theta _{m-n}\ldots \theta _{m-1}}$ , where $\theta _{m-n}\ldots \theta _{m-1}\in L_n(\theta )$ . Thus, we have that

(4.1) $$ \begin{align} \mathcal A_n(K)\subset \{A_{i_1\ldots i_n}\ :\ i_1\ldots i_n\in L_n(\theta)\}. \end{align} $$

In addition, the reciprocal contention of (4.1) is clearly valid.

Lemma 4.3 Suppose that f satisfies the separation property and $D\subset \widetilde X$ . Then, there exist $k\geq 1$ and $x_1,\ldots ,x_k\in \Lambda \cap \widetilde X$ such that

$$\begin{align*}\Lambda\subset \bigcup_{j=1}^k\left(\bigcup_{w\in L_n(\theta_j)}A_{w}\!\right)\qquad\forall\,n\geq1, \end{align*}$$

where $\theta _j$ is the itinerary of $x_j$ for every $j\in \{1,\ldots ,k\}$ . In particular, we have that ${\#\mathcal A_n=p_{\theta _1}(n)+\cdots +p_{\theta _k}(n)}$ .

We want to have a notion of the attractor size of a PCIM. For this purpose, we will compute the Hausdorff dimension of $\Lambda $ using the comparison given in Lemma 2.1. Also, we will need the following elementary inequality.

Lemma 4.4 (log-sum inequality, Theorem 2.3 in [Reference Han and Kobayashi7])

Let $k\geq 1$ , and let $a_1,\ldots ,a_k$ be nonnegative real numbers, then

$$\begin{align*}\left(\sum_{i=1}^k a_i\right)\log\!\left(\frac{1}{k}\sum_{i=1}^k a_i\right)\leq \sum_{i=1}^ka_i \log a_i. \end{align*}$$

The previous lemma will allow us to establish an upper bound for the box dimension and, therefore, for the Hausdorff dimension. With this idea, we can prove the following result.

Proof Thanks to Lemma 2.1, it is enough to prove that $\dim _B(\Lambda )=0$ . By (3.2), we have that

(4.2) $$ \begin{align} \Lambda\subset \bigcup_{A\in\mathcal A_n}A\qquad\forall\,n\geq1. \end{align} $$

Let $n\geq 1$ and $\epsilon _n:=\lambda ^n\operatorname {\mathrm {diam}}(X)$ . From item 1 of Lemma 3.3, we deduce that $\operatorname {\mathrm {diam}}(A)<\epsilon _n$ for every $A\in \mathcal A_n$ . Next, from (4.2) and Lemma 4.3, there exist $x_1,\ldots ,x_k\in \Lambda \cap \widetilde X$ such that

$$\begin{align*}\ell(\Lambda,\epsilon_n)\leq \#\mathcal A_{n}=\sum_{j=1}^k p_{\theta_j}(n), \end{align*}$$

where $\theta _j$ is the itinerary of $x_j$ for every $j\in \{1,\ldots ,k\}$ . Now, applying the log-sum inequality to the nonnegative numbers $k\,p_{\theta _1}(n),\ldots ,k\,p_{\theta _k}(n)$ , we obtain

$$\begin{align*}\log \ell(\Lambda,\epsilon_n)\leq \log\!\left(\frac{1}{k}\sum_{j=1}^k k\,p_{\theta_j}(n)\!\right)\leq \frac{\sum\limits_{j=1}^k p_{\theta_j}(n)\log \!\big(k\,p_{\theta_j}(n)\big)}{\sum\limits_{j=1}^k p_{\theta_j}(n)}\leq \sum_{j=1}^k \log \!\big(k\,p_{\theta_j}(n)\big). \end{align*}$$

Thus, we deduce that

(4.3) $$ \begin{align} \frac{\log \ell(\Lambda,\epsilon_n)}{\log(1/\epsilon_n)}\leq \sum_{j=1}^k\frac{\log \!\big(k\,p_{\theta_j}(n)\big)}{\log(1/\epsilon_n)}=\sum_{j=1}^k\frac{\log \!\big(k\,p_{\theta_j}(n)\big)}{n\log(1/\lambda)+\log(1/\operatorname{\mathrm{diam}}(X))}. \end{align} $$

By Theorem 3.1, we have that $p_{\theta _j}(n)$ is at most an affine function for n sufficiently large. Therefore,

$$\begin{align*}\frac{\log\!\big(k\,p_{\theta_j}(n)\big)}{n\log(1/\lambda)+\log(1/\operatorname{\mathrm{diam}}(X))}\;\stackrel{n\to\infty}\longrightarrow\; 0\qquad\forall\,j\in\{1,\ldots,k\}. \end{align*}$$

Thus, from (4.3), we deduce that $\liminf _{\epsilon \to 0^+} \frac {\log \ell (\Lambda ,\epsilon )}{\log (1/\epsilon )}\leq 0$ , concluding that ${\dim _B(\Lambda )=0}$ . Thus, by Lemma 2.1, we have that $\dim _{\mathcal H}(\Lambda )=0$ .

Theorem 4.5 corresponds to the first part of Theorem 1.3. Now, we are going to prove the second and final part. Recall that $\Lambda \cap \widetilde X$ is a nonempty and f-invariant set whenever f is piecewise monotonic and $D\subset \widetilde X$ . Given $n\geq 1$ and $\epsilon>0$ , we say that a set $E\subset \Lambda \cap \widetilde X$ is an $(n,\epsilon )$ -spanning set for $\Lambda \cap \widetilde X$ if, for every $x\in \Lambda \cap \widetilde X$ , there exists $y\in E$ such that

$$\begin{align*}\big|f^j(x)-f^j(y)\big|<\epsilon\qquad\forall\,j\in\{0,\ldots,n-1\}. \end{align*}$$

Thus, we define $r_n(\epsilon ,\Lambda \cap \widetilde X)$ as the smallest cardinality of every $(n,\epsilon )$ -spanning set for $\Lambda \cap \widetilde X$ with respect to f. The number $r_n(\epsilon ,K)$ is well defined when K is a compact set over which f is continuous; however, this is not the usual case for PCIMs.

Proof Let $n\geq 1$ and $\epsilon>0$ . From item 2 of Lemma 3.3, there exists $n_0\geq n$ such that

(4.4) $$ \begin{align} \operatorname{\mathrm{diam}} A<\epsilon\qquad \forall\,A\in\mathcal A_{n_0}. \end{align} $$

Next, denote by $\mathcal C(\mathcal A_{n_0},n)$ the collection of all connected components of

$$\begin{align*}A\,{\backslash}\, \big(\Delta\cup f^{-1}(\Delta)\cup\cdots\cup f^{-n+1}(\Delta)\big)\quad\text{with } A \in \mathcal A_{n_0}. \end{align*}$$

Since f is piecewise monotonic, the collection $\mathcal C(\mathcal A_{n_0},n)$ is finite. Furthermore, from (3.2), we can consider the subcollection $\mathcal C^*(\mathcal A_{n_0},n)\subset \mathcal C(\mathcal A_{n_0},n)$ of connected components that intersect $\Lambda \cap \widetilde X$ , whose union covers said set. Thus, for every $B\in \mathcal C^*(\mathcal A_{n_0},n)$ , we can take $x_B\in B\cap \Lambda \cap \widetilde X\neq \emptyset $ . It is clear that $E_{n,\epsilon }:=\{x_B\ :\ B\in \mathcal C^*(\mathcal A_{n_0},n)\}\subset \Lambda \cap \widetilde X$ is a finite $(n,\epsilon )$ -spanning set. This implies that $r_n(\epsilon ,\Lambda \cap \widetilde X)<\infty $ , concluding the proof.

Next, we establish and prove the last part of Theorem 1.3.

Proof Let $n\geq 1$ , and let $\epsilon>0$ be such that

$$\begin{align*}\epsilon<\min\!\big\{|c-c'|\ :\ c,c'\in\Delta \,\text{ such that } c\neq c' \big\}. \end{align*}$$

Consider $n_0\geq n$ such that (4.4) holds. Note that every atom of generation $n_0$ or higher has at most one element of $\Delta $ . Since f satisfies the separation property, we deduce, from (3.1) and Lemma 4.3, that

$$\begin{align*}\#\mathcal A_{n+n_0}-\#\mathcal A_{n_0}=\sum_{m=n_0}^{n_0+n-1}\big(\#\mathcal A_{m+1}-\#\mathcal A_{m}\big)\leq n(N-1). \end{align*}$$

Next, noting that $\#\mathcal C(\mathcal A_{n_0},n)=\#\mathcal A_{n+n_0}$ , we have the following inequalities:

$$\begin{align*}r_n(\epsilon,\Lambda\cap\widetilde X) \leq \#\mathcal C(\mathcal A_{n_0},n)\leq n(N-1)+\#\mathcal A_{n_0}. \end{align*}$$

Thus, we obtain that

$$\begin{align*}h_{top}\big(f|(\Lambda\cap\widetilde X)\big)\leq \lim_{\epsilon\to0}\limsup_{n\to\infty}\frac{\log \big(n(N-1)+\#\mathcal A_{n_0}\big)}{n}=0, \end{align*}$$

which concludes the proof.