Bohr chaoticity of topological dynamical systems

Abstract

We introduce the notion of Bohr chaoticity, which is a topological invariant for topological dynamical systems, and which means that the system is not orthogonal to any non-trivial weights. We prove the Bohr chaoticity for all systems which have a horseshoe and for all toral affine dynamical systems of positive entropy, some of which don’t have a horseshoe.

Appendices

7 Appendix A: Extended horseshoes have no roots

The extended horseshoe has no root. A self-contained proof of this fact will be given below for the case of one-sided horseshoe. A different proof is given for the case of two-sided horseshoe and it replies on the no existence of root for the two-sided full shift, which is known ( [16]). It seems that the first proof can not be adapted to the case of two-sided horseshoe.

1.1 7.1 No root of $T: \Lambda ^*\rightarrow \Lambda ^*$ (one-sided case)

Here we give a proof of Proposition 5.2.

The proof is based on the investigation of the maximal entropy measure of the dynamical system $(\Lambda ^*,T)$. Given a topological dynamical system (X, T) and an integer $s\ge 2$. A T-invariant measure is $T^s$-invariant. If such a measure is $T^s$-ergodic, it is T-ergodic. But in general, a T-ergodic measure is not necessarily $T^s$-ergodic. That is the case for the invariant measure $\frac{1}{2}(\delta _{\frac{1}{3}} + \delta _{\frac{2}{3}})$ of the doubling dynamics $x \rightarrow 2 x \mod 1$. The proof of Proposition 5.2 is based on the following lemma, which shows that the existence of a T-invariant and $T^s$-ergodic measure is an obstruction for $T^{ns}$ to be both 2-to-1 and surjective.

Lemma 7.1

Let (X, T) be a topological dynamical system and $s\ge 2$ an integer. Suppose that there exists a T-invariant measure which is $T^s$-ergodic. Then $T^{ns}$ can not be both exactly 2-to-1 and surjective for any integer $n\ge 1$.

Proof

Suppose that $T^{ns}:X\rightarrow X$ is 2-to-1 and surjective for some integer $n\ge 1$. Then the map $T:X\rightarrow X$ must be surjective and each point $x\in X$ has at most two pre-images by T, i.e. $1\le \#T^{-1}(x) \le 2$. This allows us to decompose X into two disjoint sets

$$\begin{aligned}A_0=\{x\in X: \#T^{-1}(x)=1\}, \qquad B_0=\{x\in X: \#T^{-1}(x)=2\}.\end{aligned}$$

We claim that these two sets are measurable. For any $\epsilon >0$, let

$$\begin{aligned}B_0^{\epsilon }=\{x\in X: \#T^{-1}(x)=2 \text { and } |y_1-y_2| \ge \epsilon \text { for distinct } y_1, y_2 \in T^{-1}(x)\}. \end{aligned}$$

Then $B_0^\epsilon$ is closed. Since $B_0=\cup _{n\ge 1} B_0^{1/n}$, it is an $F_\sigma$ set. Hence, $B_0$ is measurable which implies $A_0$ is also measurable.

For $0\le k\le ns$, let $B_{k}=T^{-k}(B_0)$. First notice that we have

$$\begin{aligned} T^{-1}(B_k) = B_{k+1}, \quad T(B_{k+1}) = B_k \qquad (0\le k\le ns -1) \end{aligned}$$

where the second equality is because of the surjectivity of T. We claim that the maps $T:B_{k+1}\rightarrow B_k$ are injective and then bijective for $1\le k\le ns-1$.

Indeed, otherwise, for some k and some $v\in B_k$ there are two distinct points $u', u''\in B_{k+1}$ such that $T(u') = T(u'')=v$. Let $w=T^{k-1}(v)$, which belongs to $B_1=T^{-1}(B_0)$. Let $z=T(w)$, which belongs to $B_0$. By the definition of $B_0$, there exists a point $w^*\in B_1$ different from w such that $T(w^*)=z$ and then there exists a point $u^*\in B_{k+1}$ different from $u', u''$ such that $T^k(u^*) = w^*$. Therefore, z has at least three $T^{k+1}$-preimages $u',u'', u^*$ and then at least three $T^{ns}$-preimages. This contradicts the fact that $T^{ns}$ is 2-to-1.

The above claim implies that $B_k\subset A_0$ for $1\le k \le ns-1$. Since $A_0$ and $B_0$ form a partition of X, we have $B_{k}\cap B_0=\emptyset$ for $1\le k \le ns-1$. Consequently, all $B_j$’s for $0\le j \le ns-1$ are disjoint. We claim that all $B_j$’s for $0\le j \le ns-1$ form a partition of X. To that end, it suffices to prove

$$\begin{aligned} A_0=\bigsqcup _{k=1}^{ns-1}B_{k}. \end{aligned}$$

(7.1)

Suppose that (7.1) is not true, which means there exists a point $x\in A_0\setminus \bigsqcup _{k=1}^{ns-1}B_{k}$. Hence $T^{k}(x)\in A_0$ for $0\le k\le ns-1$, by the definition of $B_k$ and the fact that $\{A_0, B_0\}$ is a partition of X. Therefore, the point $T^{ns-1}(x)\in A_0$ has a unique $T^{ns}$-preimage $T^{-1}(x)$, which contradicts to the assumption that $T^{ns}$ is 2-to-1.

Since $X= A_0\sqcup B_0$, we get the decomposition

$$\begin{aligned}X=\bigsqcup _{k=0}^{ns-1}B_{k} = \bigsqcup _{j=0}^{s-1} T^{-j} (X') \ \ \ \mathrm{with }\ \ X^{\prime }=\bigsqcup _{k=0}^{n-1}B_{ks}. \end{aligned}$$

By the hypothesis, there exists a T-invariant measure $\mu$ which is $T^s$-ergodic. The T-invariance of $\mu$ implies that $\mu (X^{\prime })=1/s$.

Let $A_1=T^{-1}(A_0)$. Recall that $B_1 =T^{-1}(B_0)$. Since $\{A_0, B_0\}$ is a partition of X, so is $\{A_1, B_1\}$. As $B_1\subset A_0$ which is proved above, we have $B_0\subset A_1$ and actually $B_0=A_1\setminus A_0$. By (7.1) and the definition of $A_1$, we get $A_1= \bigsqcup _{k=2}^{ns}B_{k}$. Then from $A_0\sqcup B_0 = A_1\sqcup B_1$ we get $B_0=B_{ns}$, in other words,

$$\begin{aligned} T^{-ns}(B_0)=B_0. \end{aligned}$$

Consequently, $T^{-s}(X^{\prime })=X^{\prime }$. Then, by the $T^s$-ergodicity of $\mu$, we have $\mu (X^{\prime })=0 \text { or } 1$, which contradicts to $\mu (X') =\frac{1}{s}$ with $s\ge 2$. $\square$

Now we shall prove Proposition 5.2 by contradiction. Basic properties of entropy function will be used. We refer to Walters’ book [33] (cf. Theorem 4.13, Theorem 7.5, Theorem 7.10, Theorem 8.1).

Proof

(Proof of Proposition 5.2)

Assume that $S:\Lambda ^*\rightarrow \Lambda ^*$ is a continuous map such that $S^n=T$ for some integer $n\ge 2$.

For $1\le j\le N-1$, we consider the dynamical system $(T^{j}(\Lambda ), T^{N})$. By the hypothesis, the map $T^j:\Lambda \rightarrow T^{j}(\Lambda )$ is a homeomorphism. So, the system $(T^{j}(\Lambda ), T^{N})$ is conjugate to $(\Lambda , T^{N})$ with the conjugation $T^{j}$. Hence, the topological entropy of the system $(T^{j}(\Lambda ), T^{N})$ is equal to $\log 2$. It follows that $h_\mathrm{top}(\Lambda ^*,T^{N})=\log 2$. Therefore,

$$\begin{aligned} h_\mathrm{top}(\Lambda ^*,T)=\frac{1}{N}\log 2. \end{aligned}$$

It follows that the root dynamical system $(\Lambda ^*,S)$ admits its topological entropy

$$\begin{aligned}h_\mathrm{top}(\Lambda ^*,S)=\frac{1}{nN}\log 2.\end{aligned}$$

Let $\mu$ be a maximal entropy measure of the system $(\Lambda ^*,S)$. It is also a maximal entropy measure of dynamical systems $(\Lambda ^*, T)$ and $(\Lambda ^*,T^{N})$.

Let $\mu _j = \mu |_{T^{j}(\Lambda )}$ be the restriction for $0\le j\le N-1$. By the T-invariance of $\mu$ and the disjointness of $T^{j}(\Lambda )$’s ($0\le j \le N-1$), it is easy to get

$$\begin{aligned} \mu _{j+1} = \mu _j \circ T^{-1} \quad (0\le j \le N-1) \end{aligned}$$

with the convention $\mu _N =\mu _0$. It follows that $\mu _j$ are all $T^{N}$-invariant. Let $\nu _j= N\cdot \mu _j$, which is a probability measure concentrated on $T^{j}(\Lambda )$. Since the systems $(T^{j}(\Lambda ), \nu _j, T^{N})$ are all conjugate, the measure-theoretic entropies $h_{\nu _j}(T^{j}(\Lambda ), T^{N})$ are equal and the common value is $h_{\nu _j}(\Lambda ^*, T^{N})$. From this, the relation $\mu = \frac{1}{N} \sum _{j=0}^{N-1}\nu _j$ and the affinity of entropy, we get

$$\begin{aligned} \log 2=h_\mu (\Lambda ^*, T^{N})= \frac{1}{N}\sum _{j=0}^{N-1} h_{\nu _j}(\Lambda ^*, T^{N}) = h_{\nu _0}(\Lambda ^*, T^{N}) = h_{\nu _0}(\Lambda , T^{N}). \end{aligned}$$

So, the measure $\nu _0$ is the unique maximal entropy measure of the horseshoe $(\Lambda , T^{N})$, i.e. the Bernoulli $(\frac{1}{2},\frac{1}{2})$-measure on $\{0,1\}^{\mathbb {N}}$. Since $\nu _0$ is $T^{N}$-ergodic, so are $\nu _j$’s. Now we claim that $\mu$ is T-ergodic. In fact, assume that $A\subset \Lambda ^*$ is a T-invariant set. Then $A\cap T^{j}(\Lambda )$ is $T^{N}$-invariant for every $0\le j\le N-1$. By the $T^{N}$-ergodicity of $\nu _j$, $\nu _j(A\cap T^{j}(\Lambda )) =0 \ \mathrm{or} \ 1$ for each j. But $\nu _j(A\cap T^{j}(\Lambda ))$ are equal for different j’s. Then we get $\mu (A) =0 \ \mathrm{or}\ 1$, because

$$\begin{aligned} \mu (A) = \frac{1}{N}\sum _{j=0}^{N-1} \nu _j(A\cap T^{j}(\Lambda )). \end{aligned}$$

The existence of S-invariant measure $\mu$ which is T-ergodic measure, contradicts the fact that $T^N: \Lambda ^*\rightarrow \Lambda ^*$ is 2-to-1, by Lemma 7.1. $\square$

1.2 7.2 No root of $T: \Lambda ^*\rightarrow \Lambda ^*$ (two-sided case)

Here we prove Proposition 6.2. The proof is based on the fact that the shift map $\sigma : \{0,1\}^\mathbb {Z}\rightarrow \{0,1\}^\mathbb {Z}$ has no root, which is well known (cf. [16], Corollary 18.2, p.371).

Given a two-sided horseshoe $(\Lambda ,T^{N})$ with disjoint steps, it is convenient to identify the subsystem $(\Lambda ^*,T)$ with the following system $(\{0,1\}^{\mathbb {Z}}\times \mathbb {Z}/N\mathbb {Z}, \sigma _N)$, a tower of height N, where $\sigma _N$ is defined by

$$\begin{aligned} \sigma _N(\omega ,k)=\left\{ \begin{array}{ll} (\sigma (\omega ),0), &{} \text{ if } k=N-1; \\ (\omega ,k+1), &{} \text{ otherwise. } \end{array} \right. \end{aligned}$$

(7.2)

The tower $\{0,1\}^{\mathbb {Z}}\times \mathbb {Z}/N\mathbb {Z}$ has N floors $F_i = \{0,1\}^{\mathbb {Z}}\times \{i\}$ for $0\le i <N$. We extend $F_i$ for all integers $i\ge 0$ by defining $F_i = F_{i \mod N}$. Especially $F_N = F_0$. In the following we denote $\sigma _N$ by T.

Suppose that T has a root S, i.e. $S^p=T$ for some $p\ge 2$. It is clear that S is bijective and commutes with $\sigma _N$. We claim that S permute floors, that is to say, for any i there exists a j such that $S (F_i) =F_j$. We shall prove this claim by the commutativity of S with T. Assume this claim for the moment. Then each floor $F_i$ ($0\le i <N$) is mapped back to $F_i$ by $S^N$, that is to say,

$$\begin{aligned} \forall \omega , \quad S^N(\omega , i) = (R_i\omega , i), \end{aligned}$$

(7.3)

where $R_i: \{0, 1\}^\mathbb {Z}\rightarrow \{0, 1\}^\mathbb {Z}$ is some map, which depends on i. It is easy to see that $R_i$ is continuous. Thus, on one hand, we have

$$\begin{aligned} \forall \omega , \quad S^{Np}(\omega , i) =(R_i^p\omega , i); \end{aligned}$$

on the other hand, we have

$$\begin{aligned} \forall \omega , \quad S^{Np}(\omega , i) = T^N( \omega , i)= ( \sigma \omega , i). \end{aligned}$$

It follows that $\sigma =R_i^p$, which is impossible.

Now let us prove the claim. Let $\mu$ the ergodic probability measure of the maximal entropy of T (its restriction on each floor is the symmetric Bernoulli measure). We have $\mu (F_i) =\frac{1}{N}$. Let $C_i = S(F_0)\cap F_i$, the portion of $SF_0$ contained in $F_i$, for $0\le i <N$. Suppose $0<\mu (C_i)<\mu (F_i)$ for some i. There would be a contradiction. Indeed, we have the invariance

$$\begin{aligned} T \left( \bigcup _{j=0}^{N-1} T^j (C_i)\right) = \bigcup _{j=0}^{N-1} T^j (C_i). \end{aligned}$$

This is because

$$\begin{aligned} T^j(C_i)= T^j(S(F_0)\cap F_i)= ST^j (F_0) \cap T^jF_{i} = S(F_j) \cap F_{i+j} \end{aligned}$$

and

$$\begin{aligned} \bigcup _{j=1}^{N} S(F_j) \cap F_{i+j} = \bigcup _{j=0}^{N-1} S(F_j) \cap F_{i+j}. \end{aligned}$$

The invariant set $\bigcup _{j=0}^{N-1} T^j (C_i)$ has its measure between 0 and 1 because of $0<\mu (C_i)<\frac{1}{N}$. This contradicts the ergodicity of $\mu$. We have thus proved that for every i, the measure $\mu (S (F_0) \cap F_i)$ is equal to 0 or $\mu (F_i)$. There exists one i such that $\mu (S (F_0) \cap F_i)= \mu (F_i)$, otherwise $\mu (S(F_0))=0$ so that

$$\begin{aligned} \mu (S(F_j))= \mu (ST^j (F_0))= \mu (T^j S (F_0)) = \mu (S(F_0))=0, \end{aligned}$$

which implies that $\mu$ is the null measure. There is at most one i such that $\mu (S (F_0) \cap F_i)= \mu (F_i)$, otherwise $\mu (S(F_0))\ge \frac{2}{N}$, which implies that $\mu$ has a total measure equal to at least 2. So, $S(F_0)$ and $F_i$ are equal almost everywhere. If we take into account the continuity of S, we get that the two compact sets $S(F_0)$ and $F_i$ must be equal. In place of $F_0$, we can consider any $F_j$. The same argument shows that $S(F_j)$ must be equal to some $F_k$. In this way, S defines a permutation on floors. Otherwise, under S we have a cycle

$$\begin{aligned} F_{i_0} \rightarrow F_{i_1}\rightarrow \cdots \rightarrow F_{i_{\ell -1}}\rightarrow F_{i_0} \end{aligned}$$

with $\ell <N$. Then the union U of these $\ell$ floors is S-invariant and it is also $S^p$-invariant, i.e. T-invariant, an obvious contradiction.

8 Appendix B: Any cylinder contains a horseshoe

1.1 8.1 Horseshoe with disjoint steps in any cylinder: one-sided case

Consider the one-sided full shift system $(\{0,1\}^{\mathbb {N}}, \sigma )$. For any non-empty open set $U\subset \{0,1\}^{\mathbb {N}}$, we shall show that there exists a horseshoe $(\Lambda ,\sigma ^N)$ with disjoint steps for some sufficiently large integer $N\ge 1$ such that $\Lambda \subset U$ and the map $\sigma ^{N-1}: \Lambda \rightarrow \sigma ^{N-1}(\Lambda )$ is bijective. In other word, this horseshoe $(\Lambda ,\sigma ^N)$ satisfies the conditions (i) and (ii) required by Proposition 5.2.

For a word $a_0\cdots a_{n-1}$ of length n, denote by $[a_0\cdots a_{n-1}]$ the cylinder of rank n

$$\begin{aligned}=\{y \in \{0,1\}^{\mathbb {N}}: y_i=a_i \text { for } 0\le i \le n-1\}.\end{aligned}$$

Denote the n-prefix of $x=(x_j)_{j\ge 0}$ by $x|_n$, i.e. $x|_n=x_0x_1\cdots x_{n-1}$. Hence $[x|_{n}]$ denote a cylinder of rank n. Let uv denote the concatenation $u_0u_1\cdots u_{n-1}v_0\cdots v_{m-1}$ of two words $u=u_0u_1\cdots u_{n-1}$ and $v= v_0\cdots v_{m-1}$. So, $u^{r}$ denotes $u\cdots u$ (r times). In particular, $1^r$ means $1\cdots 1$ (r times). The following lemma is a preparation for proving the above announced existence of horseshoe in a given cylinder.

Lemma 8.1

For any cylinder C in $\{0,1\}^\mathbb {N}$ of rank $M\ge 1$, there exists a sub-cylinder $C'\subset C$ of rank N with $N\ge M$ such that

• (i) $\sigma ^N (C') = \{0,1\}^\mathbb {N};$

• (ii) $C'\cap \sigma ^{n}(C')=\emptyset \ \text {for } 1\le n \le N-1.$

Proof

Assume $C=[a_0a_1\cdots a_{M-1}]$. We assume $a_0=0$, without loss of generality. Note that $\sigma ^{M}(C)=\{0,1\}^{\mathbb {N}}$. Let $n_0\ge 1$ be the minimal positive integer such that $C \subset \sigma ^{n_0}(C)$, so that

$$\begin{aligned} C\cap \sigma ^{n}(C)=\emptyset \quad \ \text {for\ all }\ 1\le n\le n_0-1. \end{aligned}$$

(8.1)

We restate these facts as follow

$$\begin{aligned}{}[0a_1\cdots a_{M-1}]\cap [a_1\cdots a_{M-1}]=\emptyset ,&\cdots , [0a_1\cdots a_{M-1}]\cap [a_{n_0-1}a_{n_0}\cdots a_{M-1}]=\emptyset ; \end{aligned}$$

(8.2)

$$\begin{aligned}{}[0a_1\cdots a_{M-1}]\subset [a_{n_0}\cdots a_{M-1}]. \end{aligned}$$

(8.3)

We have $n_0\le M$. If $n_0=M$, we are done and we can take $C'=C$. In the following, we assume $n_0<M$.

The inclusion (8.3) means

$$\begin{aligned} a_{n_0+j}=a_j \quad \text {for } 0\le j \le M-n_0-1. \end{aligned}$$

(8.4)

Let $x=\overline{a_0a_1\cdots a_{n_0-1}}$, which is $n_0$-periodic. By (8.4), $a_0a_1\cdots a_{M-1}$ is a prefix of x, so x is in C. By (8.2), $n_0$ is the minimal period of x. Define the sub-cylinder $C^{\prime }=[0a_1a_2\cdots a_{M-1}1^{n_0}]$, or more precisely

$$\begin{aligned} C^{\prime } =[(0a_1\cdots a_{n_0-1})^{q}0a_1\cdots a_{j-1}1^{n_0}] \end{aligned}$$

where $q\ge 0$ and $0\le j \le n_0-1$ are determined by $M=qn_0+j$. Now we shall check that the sub-cylinder $C^{\prime }$(of C) of rank $N=M+n_0$ has property (ii). We distinguish three cases.

Case I. $1\le n\le n_{0}-1$. Since $C^{\prime }\subset C$, we proved $C^{\prime }\cap \sigma ^{n}(C^{\prime })=\emptyset$, by (8.1).

Case II. $n_0\le n \le M-1$. Suppose $C^{\prime }\cap \sigma ^{n}(C^{\prime })\ne \emptyset$. Then $C\cap \sigma ^{n}(C^{\prime })\ne \emptyset$. Since $|\sigma ^{n}(C^{\prime })|\le M=|C|$, we get $C\subset \sigma ^{n}(C^{\prime })$. Hence, the word $a_n\cdots a_{M-1}1^{n_0}$ defining the cylinder $\sigma ^{n}(C^{\prime })$ has $1^{n_0}$ as suffix, which is a word contained in $a_0a_1\cdots a_{M-1}$ defining the cylinder C. But, on the other hand, any word of length $n_0$ contained in $a_0a_1\cdots a_{M-1}$ contains 0, a contradiction.

Case III. $M\le n \le N-1$. This case is evident because $\sigma ^{n}(C^{\prime })\subset [1]$, but $C'\subset [0]$. $\square$

Let us look at the cylinders C of rank $M=4$ and the cylinders $C'$ constructed in Lemma 8.1:

$$\begin{aligned} C=[0000],&C' = [00001^1]\\ C=[0001],&C' = C\ \ \ \ \ \ \ \ \\ C=[0010],&C' = [00101^3]\\ C=[0011],&C' = C\ \ \ \ \ \ \ \ \\ C=[0100],&C' = [01001^3]\\ C=[0101],&C' = [01011^2]\\ C=[0110],&C' = [01101^3]\\ C=[0111],&C' = C\ \ \ \ \ \ \ \ \end{aligned}$$

where the exponent represents $n_0$.

We are now ready to prove the existence of horseshoe contained in a given cylinder.

Proposition 8.2

Let $C\subset \{0,1\}^{\mathbb {N}}$ be an arbitrary cylinder. There exists a $\sigma ^{N}$-invariant closed subset $\Lambda \subset C$ for some integer $N\ge |C|$, such that

• (i) The system $(\Lambda , \sigma ^N)$ is topologically conjugate to the full shift $(\{0,1\}^{\mathbb {N}}, \sigma )$;

• (ii) The maps $\sigma ^{n}: \Lambda \rightarrow \sigma ^{n}(\Lambda ) \ (1\le n\le N-1)$ are bijections;

• (iii) The sets $\Lambda , \sigma (\Lambda ), \ldots , \sigma ^{N-1}(\Lambda )$ are disjoint.

Proof

Assume $C=[y_0y_1\cdots y_{m-1}]$ be a cylinder of rank m. We can assume $y_0=0$ without loss of generality. By Lemma 8.1, there exists an integer $n_* \ge m$ and a sub-cylinder $C^{\prime }\subset C$ of rank $n_*$ such that

$$\begin{aligned} C^{\prime }\cap \sigma ^{n}(C^{\prime })=\emptyset \ \ \text { for } 1\le n \le n_*-1. \end{aligned}$$

(8.5)

It is obvious that $\sigma ^{n_*}(C^{\prime })=\{0,1\}^\mathbb {N}$ and $\sigma ^{n_*}: C' \rightarrow \{0,1\}^\mathbb {N}$ is bijective. Assume that $C' =[y_0y_1\cdots y_{m-1} y_m \cdots y_{n_*-1}]$. Notice that $y_{n_*-1}=1$ because $C'\cap \sigma ^{n_*-1}(C')=\emptyset$. Let $y= \overline{y_0y_1\cdots y_{n_*-1}}$. This periodic point y is the unique periodic point contained in $C^{\prime }$ of exact period $n_*$, by (8.5). Let $n_1=\min \{0\le i\le n_*-1: y_i=1\}$. Since $y_0=0$ and $y_{n_*-1}=1$, we have $1\le n_1\le n_*-1$.

Define two sub-cylinders of $C'$ of rank $n_*+n_1+2$:

$$\begin{aligned} C_1= & {} [y_0y_1\cdots y_{n_*-1}10 0^{n_1}] =[0^{n_1}y_{n_1}\cdots y_{n_*-1}10 0^{n_1}]=[0^{n_1}a0^{n_1}], \\ C_2= & {} [y_0y_1\cdots y_{n_*-1}11 0^{n_1}] = [0^{n_1}y_{n_1}\cdots y_{n_*-1}11 0^{n_1}] =[0^{n_1}b0^{n_1}] \end{aligned}$$

where $a = y_{n_1}\cdots y_{n_*-1}10$ and $b = y_{n_1}\cdots y_{n_*-1}11$.

Also observe that

(iv) for $1\le n\le n_*-1$, $\sigma ^{n}(C_1\cup C_2)\cap C^{\prime }=\emptyset$ and $\sigma ^{n}$ is injective on $C_1\cup C_2$.

This follows from the relation $C_{1}\cup C_2 \subset C^{\prime }$, the disjointness (8.5) and the injectivity of $\sigma ^{n_*}: C^{\prime } \rightarrow \{0,1\}^{\mathbb {N}}$. By the definition of $n_1$, we have $C_{1}\cup C_{2}\subset [0^{n_1}1]$. Note that

$$\begin{aligned}\sigma ^{n_*}(C_1)=[100^{n_1}], \quad \sigma ^{n_*+1}(C_1)=[0^{n_1+1}]; \quad \sigma ^{n_*}(C_2)=[110^{n_1}], \quad \sigma ^{n_*+1}(C_2)=[10^{n_1}]\end{aligned}$$

which are all disjoint from $[0^{n_1}1]$. Hence,

(v) for $n=n_*$ or $n_*+1$, $\sigma ^{n}(C_1\cup C_2)\cap (C_{1}\cup C_{2})=\emptyset$ and $\sigma ^n$ is injective on $C_1\cup C_2$.

Now take $N=n_*+2$ and define

$$\begin{aligned} \Lambda =\{x\in \{0,1\}^{\mathbb {N}}: \forall k\ge 0, \sigma ^{kN}(x) \in C_{1}\cup C_{2}\}. \end{aligned}$$

Observe that both the words defining $C_1$ and $C_2$ have $0^{n_1}$ as their prefix and as well as their suffix. Therefore $\Lambda$ can be identified with the symbolic space $\{0^{n_1}a, 0^{n_1} b\}^\mathbb {N}$ and $(\Lambda , \sigma ^N)$ is conjugate to the shift map on $\{0^{n_1}a, 0^{n_1} b\}^\mathbb {N}$. This is the property (i) of $(\Lambda , \sigma ^N)$. The required properties (ii) and (iii) follow from (iv) and (v).

$\square$

1.2 8.2 Horseshoe with disjoint steps in any cylinder: two-sided case

Consider the full shift system $(\{0,1\}^{\mathbb {Z}}, \sigma )$. For any non-empty cylinder $C\subset \{0,1\}^{\mathbb {Z}}$, we shall show that there exists a horseshoe $(\Lambda ,\sigma ^N)$ with disjoint steps for some sufficiently large integer $N\ge 1$, where $\Lambda \subset C$. We start with the following preparative lemma, the counterpart of Lemma 8.1.

Lemma 8.3

For any cylinder C in $\{0,1\}^\mathbb {Z}$ of rank $M\ge 1$, there exists a sub-cylinder $C'\subset C$ of rank N with $N\ge M$ such that

• (i) $\sigma ^N (C') \cap C'\ne \emptyset ;$

• (ii) $C'\cap \sigma ^{n}(C')=\emptyset \ \text {for } 1\le n \le N-1.$

Proof

The proof is almost the same as that of Lemma 8.1. We assume $C=[a_0a_1\cdots a_{M-1}]$, and $a_0=0$ without loss of generality. Note that $C\cap \sigma ^{M}(C)\ne \emptyset$. Let $n_0\ge 1$ be the minimal positive integer such that $C\cap \sigma ^{n_0}(C)\ne \emptyset$, so that

$$\begin{aligned} C\cap \sigma ^{n}(C)=\emptyset \quad \ \text {for\ all }\ 1\le n\le n_0-1. \end{aligned}$$

(8.6)

We restate these facts as follow

$$\begin{aligned}{}[0a_1\cdots a_{M-1}]_0\cap [a_1\cdots a_{M-1}]_0=\emptyset ,&\cdots , [0a_1\cdots a_{M-1}]_0\cap [a_{n_0-1}a_{n_0}\cdots a_{M-1}]_0=\emptyset ; \end{aligned}$$

(8.7)

$$\begin{aligned}{}[0a_1\cdots a_{M-1}]_0\subset [a_{n_0}\cdots a_{M-1}]_0. \end{aligned}$$

(8.8)

The rest of the proof is the same if we replace cylinders of the for $[*]$ by $[*]_0$ where $*$ represents a word.

$\square$

We are now ready to prove the existence of horseshoe contained in a given cylinder.

Proposition 8.4

Let $C\subset \{0,1\}^{\mathbb {N}}$ be an arbitrary cylinder. There exists a $\sigma ^{N}$-invariant closed subset $\Lambda \subset C$ for some integer $N\ge |C|$, such that

• (i) The system $(\Lambda , \sigma ^N)$ is topologically conjugate to the full shift $(\{0,1\}^{\mathbb {N}}, \sigma )$;

• (ii) The sets $\Lambda , \sigma (\Lambda ), \ldots , \sigma ^{N-1}(\Lambda )$ are disjoint.

Proof

It is exactly the same proof as that of Proposition 8.2, if we replace cylinders of the form $[*]$ by $[*]_0$.

$\square$