Proof. Let R denote the right-hand side of the equation. $\operatorname {CR}(f)\subset R$ is clearly true. Let us prove $R\subset \operatorname {CR}(f)$ . Note that $\pi _n^{n+1}(\operatorname {CR}(f_{n+1}))\subset \operatorname {CR}(f_n)$ for every $n\ge 1$ . Let $Y_n=\operatorname {CR}(f_n)$ , $g_n=(f_n)|_{Y_n}\colon Y_n\to Y_n$ , and $\tilde {\pi }_n^{n+1}=(\pi _n^{n+1})|_{Y_{n+1}}\colon Y_{n+1}\to Y_n$ for each $n\ge 1$ . Consider the inverse sequence of equivariant maps

$$ \begin{align*} \tilde{\pi}=(\tilde{\pi}_n^{n+1}\colon(Y_{n+1},g_{n+1})\to(Y_n,g_n))_{n\ge1} \end{align*} $$

and let $(Y,g)=\lim _{\tilde {\pi }}(Y_n,g_n)$ . Since $g_n$ is chain recurrent for all $n\ge 1$ , by Lemma 2.3, g is chain recurrent. On the other hand, R is a closed f-invariant subset of X and satisfies $Y=R$ . The inclusion $i\colon Y\to R$ gives a topological conjugacy $i\colon (Y,g)\to (R,f|_R)$ , and so $f|_R\colon R\to R$ is chain recurrent, which clearly implies $R\subset \operatorname {CR}(f)$ ; therefore, the lemma has been proved.

Proof. For any $C_{\ast }=(C_n)_{n\ge 1}\in \mathcal {C}_\pi $ , $[C_{\ast }]$ is a closed f-invariant subset of X. We prove that $f|_{[C_{\ast }]}\colon [C_{\ast }]\to [C_{\ast }]$ is chain transitive. For each $n\ge 1$ , let $g_n=f|_{C_n}\colon C_n\to C_n$ and let

$$ \begin{align*} \tilde{\pi}_n^{n+1}=(\pi_n^{n+1})|_{C_{n+1}}\colon C_{n+1}\to C_n. \end{align*} $$

Consider the inverse sequence of equivariant maps

$$ \begin{align*} \tilde{\pi}=(\tilde{\pi}_n^{n+1}\colon(C_{n+1},g_{n+1})\to(C_n,g_n))_{n\ge1} \end{align*} $$

and let $(Y,g)=\lim _{\tilde {\pi }}(C_n,g_n)$ . Since $g_n$ is chain transitive for all $n\ge 1$ , by Lemma 2.3, g is chain transitive. On the other hand, we have $Y=[C_{\ast }]$ . The inclusion $i\colon Y\to [C_{\ast }]$ gives a topological conjugacy $i\colon (Y,g)\to ([C_{\ast }],f|_{[C_{\ast }]})$ , which implies that $f|_{[C_{\ast }]}$ is chain transitive.

Given any $C_{\ast }=(C_n)_{n\ge 1}\in \mathcal {C}_\pi $ , from what is shown above, there is $C\in \mathcal {C}(f)$ such that $[C_{\ast }]\subset C$ . Fix $x=(x_n)_{n\ge 1}\in [C_{\ast }]$ . Then, for every $y=(y_n)_{n\ge 1}\in C$ , we easily see that $\{x_n,y_n\}\in C_n$ for all $n\ge 1$ ; therefore, $y\in [C_{\ast }]$ . This implies $C\subset [C_{\ast }]$ and so $[C_{\ast }]=C$ , proving

$$ \begin{align*} \{[C_{\ast}]\colon C_{\ast}\in\mathcal{C}_\pi\}\subset\mathcal{C}(f). \end{align*} $$

To prove

$$ \begin{align*} \mathcal{C}(f)\subset\{[C_{\ast}]\colon C_{\ast}\in\mathcal{C}_\pi\}, \end{align*} $$

for any $C\in \mathcal {C}(f)$ , fix $x=(x_n)_{n\ge 1}\in C$ , and take $C_n\in \mathcal {C}(f_n)$ with $x_n\in C_n$ for each $n\ge 1$ . Then $C_{\ast }=(C_n)_{n\ge 1}\in \mathcal {C}_\pi $ and $x\in [C_{\ast }]\subset C$ . Similarly to the argument above, we obtain $C=[C_{\ast }]$ , completing the proof.

Proof. Let $D_{\ast }=(D_n)_{n\ge 1}\in \mathcal {D}_\pi $ and let $x=(x_n)_{n\ge 1},y=(y_n)_{n\ge 1}\in [D_{\ast }]$ . We prove that $(x,y)\in X^2$ is chain proximal for f. Fix any $N>0$ and $\delta>0$ . Since $\{x_{N+1},y_{N+1}\}\subset D_{N+1}\in \mathcal {D}(f_{N+1})$ , $(x_{N+1},y_{N+1})\in X_{N+1}^2$ is chain proximal for $f_{N+1}$ , implying that there is a pair

$$ \begin{align*} ((x_{N+1}^{(i)})_{i=0}^{k},(y_{N+1}^{(i)})_{i=0}^{k}) \end{align*} $$

of $\delta $ -chains of $f_{N+1}$ such that $(x_{N+1}^{(0)},y_{N+1}^{(0)})=(x_{N+1},y_{N+1})$ and $x_{N+1}^{(k)}=y_{N+1}^{(k)}$ . Let $(z^{(0)},w^{(0)})=(x,y)\in X^2$ and note that

$$ \begin{align*} (z_n^{(0)},w_n^{(0)})=(x_n,y_n)=(\pi_n^{N+1}(x_{N+1}),\pi_n^{N+1}(y_{N+1}))=(\pi_n^{N+1}(x_{N+1}^{(0)}),\pi_n^{N+1}(y_{N+1}^{(0)})) \end{align*} $$

for every $1\le n\le N$ . For each $0<i\le k$ , since

$$ \begin{align*} \{\pi_N^{N+1}(x_{N+1}^{(i)}),\pi_N^{N+1}(y_{N+1}^{(i)})\}\subset\pi_N^{N+1}(X_{N+1})=\hat{X}_N\quad \end{align*} $$

by $\operatorname {MLC}(1)$ (see Lemma 2.1), there are $(z^{(i)},w^{(i)})\in X^2$ , $0<i \le k-1$ , and $z^{(k)}=w^{(k)}\in X$ such that

$$ \begin{align*} (z_n^{(i)},w_n^{(i)})=(\pi_n^{N+1}(x_{N+1}^{(i)}),\pi_n^{N+1}(y_{N+1}^{(i)})) \end{align*} $$

and also

$$ \begin{align*} z_n^{(k)}=w_n^{(k)}=\pi_n^{N+1}(x_{N+1}^{(k)})=\pi_n^{N+1}(y_{N+1}^{(k)}) \end{align*} $$

for every $1\le n\le N$ . Let $d_n$ , $n\ge 1$ , be the metric on $X_n$ . For any $0\le i\le k-1$ and $1\le n\le N$ , we have

$$ \begin{align*} \begin{aligned} d_n(f(z^{(i)})_n,z_n^{(i+1)})&=d_n(f_n(z_n^{(i)}),z_n^{(i+1)})\\ &=d_n(f_n(\pi_n^{N+1}(x_{N+1}^{(i)})),\pi_n^{N+1}(x_{N+1}^{(i+1)}))\\ &=d_n(\pi_n^{N+1}(f_{N+1}(x_{N+1}^{(i)})),\pi_n^{N+1}(x_{N+1}^{(i+1)})) \end{aligned} \end{align*} $$

with $d_{N+1}(f_{N+1}(x_{N+1}^{(i)}),x_{N+1}^{(i+1)})\le \delta $ , and similarly,

$$ \begin{align*} d_n(f(w^{(i)})_n,w_n^{(i+1)})=d_n(\pi_n^{N+1}(f_{N+1}(y_{N+1}^{(i)})),\pi_n^{N+1}(y_{N+1}^{(i+1)})) \end{align*} $$

with $d_{N+1}(f_{N+1}(y_{N+1}^{(i)}),y_{N+1}^{(i+1)})\le \delta $ . Therefore, for every $\epsilon>0$ , if N is large enough, and then $\delta $ is sufficiently small,

$$ \begin{align*} ((z^{(i)})_{i=0}^{k},(w^{(i)})_{i=0}^{k}) \end{align*} $$

is a pair of $\epsilon $ -chains of f with $(z^{(0)},w^{(0)})=(x,y)$ and $z^{(k)}=w^{(k)}$ , proving that $(x,y)\in X^2$ is chain proximal for f.

Given any $D_{\ast }=(D_n)_{n\ge 1}\in \mathcal {D}_\pi $ , from what is shown above, we have $[D_{\ast }]\subset D$ for some $D\in \mathcal {D}(f)$ . The rest of the proof is identical to that of Lemma 3.2.

Proof. For any $n\ge 1$ and $q\in X_{n+1}$ , since $\pi _n^{n+1}(q)\in \pi _n^{n+1}(X_{n+1})=\hat {X}_n$ by $\operatorname {MLC}(1)$ of $\pi $ (see Lemma 2.1), we have $x_n=\pi _n^{n+1}(q)$ for some $x=(x_j)_{j\ge 1}\in X$ . Since $\pi _n^{n+1}(q)=x_n=\pi _n^{n+2}(x_{n+2})$ , by property (2), we obtain $\pi _n^{n+1}(q)\in \pi _n^{n+2}(Y_{n+2})$ , implying $\pi _n^{n+1}(X_{n+1})\subset \pi _n^{n+2}(Y_{n+2})$ . Then

$$ \begin{align*} \pi_n^{n+1}(Y_{n+1})\subset\pi_n^{n+1}(X_{n+1})\subset\pi_n^{n+2}(Y_{n+2})\subset\pi_n^{n+1}(Y_{n+1}), \end{align*} $$

therefore $\pi _n^{n+1}(Y_{n+1})=\pi _n^{n+2}(Y_{n+2})$ . Since $n\ge 1$ is arbitrary, $\tilde {\pi }$ satisfies $\operatorname {MLC}(1)$ .

Proof. Let $n\ge 1$ . Since $Y_{n+1}=\overline {\operatorname {Per}(f_{n+1})}$ and $\pi _n^{n+1}(\operatorname {Per}(f_{n+1}))\subset \operatorname {Per}(f_n)$ ,

$$ \begin{align*} \pi_n^{n+1}(Y_{n+1})=\pi_n^{n+1}(\overline{\operatorname{Per}(f_{n+1})})\subset\overline{\pi_n^{n+1}(\operatorname{Per}(f_{n+1}))}\subset\overline{\pi_n^{n+1}(Y_{n+1})\cap \operatorname{Per}(f_n)}. \end{align*} $$

By $\operatorname {MLC}(1)$ of $\pi $ , $\pi _n^{n+1}(Y_{n+1})\subset \pi _n^{n+1}(X_{n+1})=\pi _n^{n+2}(X_{n+2})$ ; therefore, for any $p\in \pi _n^{n+1}(Y_{n+1})\cap \operatorname {Per}(f_n)$ , there is $q\in X_{n+2}$ such that $p=\pi _n^{n+2}(q)$ . Then there is $r\in Y_{n+2}$ such that

$$ \begin{align*} \lim_{k\to\infty}d_{n+2}(f_{n+2}^k(q),f_{n+2}^k(r))=0, \end{align*} $$

implying

$$ \begin{align*} \begin{aligned} \lim_{k\to\infty}d_n(f_n^k(p),f_n^k(\pi_{n}^{n+2}(r)))&=\lim_{k\to\infty}d_n(f_n^k(\pi_{n}^{n+2}(q)),f_n^k(\pi_{n}^{n+2}(r)))\\ &=\lim_{k\to\infty}d_n(\pi_{n}^{n+2}(f_{n+2}^k(q)),\pi_{n}^{n+2}(f_{n+2}^k(r)))\\ &=0, \end{aligned} \end{align*} $$

where $d_n$ , $d_{n+2}$ are the metrics on $X_n$ , $X_{n+2}$ . Note that $\pi _{n}^{n+2}(r)\in \pi _n^{n+2}(Y_{n+2})$ . From $p\in \operatorname {Per}(f_n)$ and the $f_n$ -invariance of $\pi _n^{n+2}(Y_{n+2})$ , it follows that $p\in \overline {\pi _n^{n+2}(Y_{n+2})}=\pi _n^{n+2}(Y_{n+2})$ . Since $p\in \pi _n^{n+1}(Y_{n+1})\cap \operatorname {Per}(f_n)$ is arbitrary, we obtain

$$ \begin{align*} \pi_n^{n+1}(Y_{n+1})\cap \operatorname{Per}(f_n)\subset\pi_n^{n+2}(Y_{n+2}) \end{align*} $$

and so

$$ \begin{align*} \pi_n^{n+1}(Y_{n+1})\subset\overline{\pi_n^{n+1}(Y_{n+1})\cap \operatorname{Per} (f_n)} \subset\overline{\pi_n^{n+2}(Y_{n+2})}=\pi_n^{n+2}(Y_{n+2}). \end{align*} $$

Thus, $\pi _n^{n+1}(Y_{n+1})=\pi _n^{n+2}(Y_{n+2})$ , proving the lemma.

Proof of Lemma 4.1

By Lemmas 1.2 and 2.4, we may assume $(X,f)=\lim _{\pi }(X_n,f_n)$ , where $(X_n,f_n)$ , $n\ge 1$ , are SFTs, and

$$ \begin{align*} \pi=(\pi_n^{n+1}\colon(X_{n+1},f_{n+1})\to(X_n,f_n))_{n\ge1} \end{align*} $$

is an inverse sequence of equivariant maps with $\operatorname {MLC}(1)$ . Let $Y_n=\operatorname {CR}(f_n)$ , $g_n=(f_n)|_{Y_n}\colon Y_n\to Y_n$ , and $\tilde {\pi }_n^{n+1}=(\pi _n^{n+1})|_{Y_{n+1}}\colon Y_{n+1}\to Y_n$ for every $n\ge 1$ . Then $(Y_n,g_n)$ , $n\ge 1$ , are chain recurrent SFTs, and by Lemma 4.2,

$$ \begin{align*} \tilde{\pi}=(\tilde{\pi}_n^{n+1}\colon(Y_{n+1},g_{n+1})\to(Y_n,g_n))_{n\ge1} \end{align*} $$

satisfies $\operatorname {MLC}(1)$ . Letting $(Y,g)=\lim _{\tilde {\pi }}(Y_n,g_n)$ and

$$ \begin{align*} R=\{x=(x_n)_{n\ge1}\in X\colon x_n\in \operatorname{CR}(f_n),\text{ for all } n\ge 1\}, \end{align*} $$

we have $R=\operatorname {CR}(f)$ by Lemma 3.1, and as in the proof of Lemma 3.1, the inclusion $i\colon Y\to \operatorname {CR}(f)$ is a topological conjugacy $i\colon (Y,g)\to (\operatorname {CR}(f),f|_{\operatorname {CR}(f)})$ . By this, again without loss of generality, we may assume that f and $f_n$ , $n\ge 1$ , are chain recurrent.

Since $h_{\mathrm {top}}(f)>0$ , there is $C^\dagger \in \mathcal {C}(f)$ such that $h_{\mathrm {top}}(f|_{C^\dagger })>0$ . A proof of this fact is as follows. By the variational principle, there is an ergodic f-invariant Borel probability measure $\mu $ on X such that the measure-theoretical entropy $h_\mu (f)$ is positive. Since $f|_{\mathrm {supp}(\mu )}\colon \mathrm {supp}(\mu )\to \mathrm {supp}(\mu )$ , the restriction of f to the support of $\mu $ , is transitive, there is $C^\dagger \in \mathcal {C}(f)$ such that $\mathrm {supp}(\mu )\subset C^\dagger $ . By the variational principle again, we obtain

$$ \begin{align*} h_{\mathrm{top}}(f|_{C^\dagger})\ge h_{\mathrm{top}}(f|_{\mathrm{supp}(\mu)})\ge h_\mu(f|_{\mathrm{supp}(\mu)})=h_\mu(f)>0. \end{align*} $$

Then, by Lemma 3.2, there is $C_{\ast }=(C_n)_{n\ge 1}\in \mathcal {C}_\pi $ such that $C^\dagger =[C_{\ast }]$ . Letting $\Gamma =\prod _{n\ge 1}\pi _n^{n+1}(C_{n+1})$ , a closed F-invariant subset of $\prod _{n\ge 1}X_n$ where $F=\prod _{n\ge 1}f_n$ is the product map, since $C^\dagger \subset \Gamma $ , we have

$$ \begin{align*} 0<h_{\mathrm{top}}(f|_{C^\dagger})=h_{\mathrm{top}}(F|_{C^\dagger})&\le h_{\mathrm{top}}(F|_\Gamma)\\ &=h_{\mathrm{top}}\bigg(\prod_{n\ge1}(f_n)|_{\pi_n^{n+1}(C_{n+1})}\bigg)=\sum_{n\ge1}h_{\mathrm{top}}((f_n)|_{\pi_n^{n+1}(C_{n+1})}) \end{align*} $$

implying $h_{\mathrm {top}}((f_n)|_{\pi _n^{n+1}(C_{n+1})})>0$ for some $n\ge 1$ .

Note that $\mathcal {C}(f_m)$ , $m\ge 1$ , are finite sets, and for any $m\ge 1$ and $D\in \mathcal {C}(f_m)$ , $(f_m)|_{D}\colon D\to D$ is transitive (see §2.4). Let us prove the following claim.

Proof of the claim

Step 1. Let $C^{\prime }_n=C_n$ . Take $D_{n+1}\in \mathcal {C}(f_{n+1})$ such that

(P1) $$ \begin{align} \pi_n^{n+1}(C_{n+1})\subset\pi_n^{n+1}(D_{n+1})\subset C^{\prime}_n \end{align} $$

and $\pi _n^{n+1}(D_{n+1})$ is maximal among

$$ \begin{align*} \{\pi_n^{n+1}(E_{n+1})\colon E_{n+1}\in\mathcal{C}(f_{n+1}),\pi_n^{n+1}(C_{n+1})\subset\pi_n^{n+1}(E_{n+1})\subset C^{\prime}_n\} \end{align*} $$

with respect to the inclusion relation.

Step 2. Note that $(f_n)|_{\pi _n^{n+1}(D_{n+1})}$ is transitive, and take a transitive point $p_1\in \pi _n^{n+1}(D_{n+1})$ , that is, $\pi _n^{n+1}(D_{n+1})=\omega (p_1,f_n)$ , the $\omega $ -limit set. Since

$$ \begin{align*} p_1\in\pi_n^{n+1}(D_{n+1})\subset\pi_n^{n+1}(X_{n+1})=\pi_n^{n+2}(X_{n+2}), \end{align*} $$

we have $p_1=\pi _n^{n+2}(q_1)$ for some $q_1\in X_{n+2}$ . Take $C^{\prime }_{n+1}\in \mathcal {C}(f_{n+1})$ with $\pi _{n+1}^{n+2}(q_1) \in C^{\prime }_{n+1}$ . Then choose $D_{n+2}\in \mathcal {C}(f_{n+2})$ such that

(P2) $$ \begin{align} \pi_{n+1}^{n+2}(q_1)\in\pi_{n+1}^{n+2}(D_{n+2})\subset C^{\prime}_{n+1} \end{align} $$

and $\pi _{n+1}^{n+2}(D_{n+2})$ is maximal among

$$ \begin{align*} \{\pi_{n+1}^{n+2}(E_{n+2})\colon E_{n+2}\in\mathcal{C}(f_{n+2}),\pi_{n+1}^{n+2}(q_1)\in\pi_{n+1}^{n+2}(E_{n+2})\subset C^{\prime}_{n+1}\} \end{align*} $$

with respect to the inclusion relation. By (P2), we have

$$ \begin{align*} p_1\in\pi_n^{n+2}(D_{n+2})\subset\pi_n^{n+1}(C^{\prime}_{n+1}), \end{align*} $$

implying

$$ \begin{align*} \pi_n^{n+1}(D_{n+1})\subset\pi_n^{n+2}(D_{n+2})\subset\pi_n^{n+1}(C^{\prime}_{n+1}) \end{align*} $$

since $\pi _n^{n+1}(D_{n+1})=\omega (p_1,f_n)$ , and $\pi _n^{n+2}(D_{n+2})$ is $f_n$ -invariant. By (P1) and $p_1\in \pi _n^{n+1}(D_{n+1})$ , we see that $\pi _n^{n+1}(C_{n+1})\subset \pi _n^{n+1}(D_{n+1})$ and $p_1\in C^{\prime }_n\cap \pi _n^{n+1}(C^{\prime }_{n+1})$ ; therefore,

$$ \begin{align*} \pi_n^{n+1}(C_{n+1})\subset\pi_n^{n+1}(D_{n+1})\subset\pi_n^{n+2}(D_{n+2})\subset\pi_n^{n+1}(C^{\prime}_{n+1})\subset C^{\prime}_n. \end{align*} $$

By the maximality of $\pi _n^{n+1}(D_{n+1})$ in Step 1, we obtain

(Q1) $$ \begin{align} \pi_n^{n+1}(D_{n+1})=\pi_n^{n+2}(D_{n+2})=\pi_n^{n+1}(C^{\prime}_{n+1}). \end{align} $$

Step 3. Note that $(f_{n+1})|_{\pi _{n+1}^{n+2}(D_{n+2})}$ is transitive, and take a transitive point $p_2\in \pi _{n+1}^{n+2}(D_{n+2})$ , that is, $\pi _{n+1}^{n+2}(D_{n+2})=\omega (p_2,f_{n+1})$ . Since

$$ \begin{align*} p_2\in\pi_{n+1}^{n+2}(D_{n+2})\subset\pi_{n+1}^{n+2}(X_{n+2})=\pi_{n+1}^{n+3}(X_{n+3}), \end{align*} $$

we have $p_2=\pi _{n+1}^{n+3}(q_2)$ for some $q_2\in X_{n+3}$ . Take $C^{\prime }_{n+2}\in \mathcal {C}(f_{n+2})$ with $\pi _{n+2}^{n+3}(q_2)\in C^{\prime }_{n+2}$ . Then choose $D_{n+3}\in \mathcal {C}(f_{n+3})$ such that

(P3) $$ \begin{align} \pi_{n+2}^{n+3}(q_2)\in\pi_{n+2}^{n+3}(D_{n+3})\subset C^{\prime}_{n+2} \end{align} $$

and $\pi _{n+2}^{n+3}(D_{n+3})$ is maximal among

$$ \begin{align*} \{\pi_{n+2}^{n+3}(E_{n+3})\colon E_{n+3}\in\mathcal{C}(f_{n+3}),\pi_{n+2}^{n+3}(q_2)\in\pi_{n+2}^{n+3}(E_{n+3})\subset C^{\prime}_{n+2}\} \end{align*} $$

with respect to the inclusion relation. By (P3), we have

$$ \begin{align*} p_2\in\pi_{n+1}^{n+3}(D_{n+3})\subset\pi_{n+1}^{n+2}(C^{\prime}_{n+2}), \end{align*} $$

implying

$$ \begin{align*} \pi_{n+1}^{n+2}(D_{n+2})\subset\pi_{n+1}^{n+3}(D_{n+3})\subset\pi_{n+1}^{n+2}(C^{\prime}_{n+2}) \end{align*} $$

since $\pi _{n+1}^{n+2}(D_{n+2})=\omega (p_2,f_{n+1})$ , and $\pi _{n+1}^{n+3}(D_{n+3})$ is $f_{n+1}$ -invariant. By (P2) and $p_2\in \pi _{n+1}^{n+2}(D_{n+2})$ , we see that $\pi _{n+1}^{n+2}(q_1)\in \pi _{n+1}^{n+2}(D_{n+2})$ and $p_2\in C^{\prime }_{n+1}\cap \pi _{n+1}^{n+2}(C^{\prime }_{n+2})$ ; therefore,

$$ \begin{align*} \pi_{n+1}^{n+2}(q_1)\in\pi_{n+1}^{n+2}(D_{n+2})\subset\pi_{n+1}^{n+3}(D_{n+3})\subset\pi_{n+1}^{n+2}(C^{\prime}_{n+2})\subset C^{\prime}_{n+1}. \end{align*} $$

By the maximality of $\pi _{n+1}^{n+2}(D_{n+2})$ in Step 2, we obtain

(Q2) $$ \begin{align} \pi_{n+1}^{n+2}(D_{n+2})=\pi_{n+1}^{n+3}(D_{n+3})=\pi_{n+1}^{n+2}(C^{\prime}_{n+2}). \end{align} $$

Assertions (Q1) and (Q2) yield $\pi _n^{n+1}(C^{\prime }_{n+1})=\pi _n^{n+2}(C^{\prime }_{n+2})$ .

Step 4. Note that $(f_{n+2})|_{\pi _{n+2}^{n+3}(D_{n+3})}$ is transitive, and take a transitive point $p_3\in \pi _{n+2}^{n+3}(D_{n+3})$ , that is, $\pi _{n+2}^{n+3}(D_{n+3})=\omega (p_3,f_{n+2})$ . Since

$$ \begin{align*} p_3\in\pi_{n+2}^{n+3}(D_{n+3})\subset\pi_{n+2}^{n+3}(X_{n+3})=\pi_{n+2}^{n+4}(X_{n+4}), \end{align*} $$

we have $p_3=\pi _{n+2}^{n+4}(q_3)$ for some $q_3\in X_{n+4}$ . Take $C^{\prime }_{n+3}\in \mathcal {C}(f_{n+3})$ with $\pi _{n+3}^{n+4}(q_3)\in C^{\prime }_{n+3}$ . Then choose $D_{n+4}\in \mathcal {C}(f_{n+4})$ such that

(P4) $$ \begin{align} \pi_{n+3}^{n+4}(q_3)\in\pi_{n+3}^{n+4}(D_{n+4})\subset C^{\prime}_{n+3} \end{align} $$

and $\pi _{n+3}^{n+4}(D_{n+4})$ is maximal among

$$ \begin{align*} \{\pi_{n+3}^{n+4}(E_{n+4})\colon E_{n+4}\in\mathcal{C}(f_{n+4}),\pi_{n+3}^{n+4}(q_3)\in\pi_{n+3}^{n+4}(E_{n+4})\subset C^{\prime}_{n+3}\} \end{align*} $$

with respect to the inclusion relation. By (P4), we have

$$ \begin{align*} p_3\in\pi_{n+2}^{n+4}(D_{n+4})\subset\pi_{n+2}^{n+3}(C^{\prime}_{n+3}), \end{align*} $$

implying

$$ \begin{align*} \pi_{n+2}^{n+3}(D_{n+3})\subset\pi_{n+2}^{n+4}(D_{n+4})\subset\pi_{n+2}^{n+3}(C^{\prime}_{n+3}) \end{align*} $$

since $\pi _{n+2}^{n+3}(D_{n+3})=\omega (p_3,f_{n+2})$ , and $\pi _{n+2}^{n+4}(D_{n+4})$ is $f_{n+2}$ -invariant. By (P3) and $p_3\in \pi _{n+2}^{n+3}(D_{n+3})$ , we see that $\pi _{n+2}^{n+3}(q_2)\in \pi _{n+2}^{n+3}(D_{n+3})$ and $p_3\in C^{\prime }_{n+2}\cap \pi _{n+2}^{n+3}(C^{\prime }_{n+3})$ ; therefore,

$$ \begin{align*} \pi_{n+2}^{n+3}(q_2)\in\pi_{n+2}^{n+3}(D_{n+3})\subset\pi_{n+2}^{n+4}(D_{n+4})\subset\pi_{n+2}^{n+3}(C^{\prime}_{n+3})\subset C^{\prime}_{n+2}. \end{align*} $$

By the maximality of $\pi _{n+2}^{n+3}(D_{n+3})$ in Step 3, we obtain

(Q3) $$ \begin{align} \pi_{n+2}^{n+3}(D_{n+3})=\pi_{n+2}^{n+4}(D_{n+4})=\pi_{n+2}^{n+3}(C^{\prime}_{n+3}). \end{align} $$

Assertions (Q2) and (Q3) yield $\pi _{n+1}^{n+2}(C^{\prime }_{n+2})=\pi _{n+1}^{n+3}(C^{\prime }_{n+3})$ .

Continuing inductively, we obtain a sequence $C^{\prime }_{\ast }=(C^{\prime }_m)_{m\ge n}\in \prod _{m\ge n}\mathcal {C}(f_m)$ . Then properties (1) and (2) are ensured in Steps 1 and 2. For any $k\ge 0$ , $\pi _{n+k}^{n+k+1}(C^{\prime }_{n+k+1})\subset C^{\prime }_{n+k}$ and $\pi _{n+k}^{n+k+1}(C^{\prime }_{n+k+1})=\pi _{n+k}^{n+k+2}(C^{\prime }_{n+k+2})$ are established in Steps $k+2$ and $k+3$ , respectively. Thus, $C^{\prime }_{\ast }$ satisfies the required properties, and so the claim has been proved.

We continue the proof of Lemma 4.1. Define $C^{\prime \prime }_{\ast }=(C^{\prime \prime }_j)_{j\ge 1}\in \prod _{j\ge 1}\mathcal {C}(f_j)$ by

$$ \begin{align*} C^{\prime\prime}_j= \begin{cases} C_j&\text{if } 1\le j<n,\\ C^{\prime}_j&\text{if } n\le j. \end{cases} \end{align*} $$

By properties (1) and (3) in the claim, we see that $C^{\prime \prime }_{\ast }\in \mathcal {C}_\pi $ . By Lemma 3.2, letting $C=[C^{\prime \prime }_{\ast }]$ , we obtain $C\in \mathcal {C}(f)$ . Let

$$ \begin{align*} \pi''=((\pi_j^{j+1})|_{C^{\prime\prime}_{j+1}}\colon(C^{\prime\prime}_{j+1},(f_{j+1})|_{C^{\prime\prime}_{j+1}})\to(C^{\prime\prime}_j,(f_j)|_{C^{\prime\prime}_j}))_{j\ge1} \end{align*} $$

and

$$ \begin{align*} \pi'=((\pi_m^{m+1})|_{C^{\prime}_{m+1}}\colon(C^{\prime}_{m+1},(f_{m+1})|_{C^{\prime}_{m+1}})\to(C^{\prime}_m,(f_m)|_{C^{\prime}_m}))_{m\ge n}. \end{align*} $$

Then $(C,f|_C)$ (respectively, $\lim _{\pi ''}(C^{\prime \prime }_j,(f_j)|_{C^{\prime \prime }_j})$ ) is topologically conjugate to

$$ \begin{align*} \lim_{\pi''}(C^{\prime\prime}_j,(f_j)|_{C^{\prime\prime}_j}) \end{align*} $$

(respectively, $\lim _{\pi '}(C^{\prime }_m,(f_m)|_{C^{\prime }_m})$ ), so $(C,f|_C)$ is topologically conjugate to

$$ \begin{align*} \lim_{\pi'}(C^{\prime}_m,(f_m)|_{C^{\prime}_m}). \end{align*} $$

Let $(Y,g)=\lim _{\pi '}(C^{\prime }_m,(f_m)|_{C^{\prime }_m})$ . By property (4) in the claim, $\pi '$ satisfies $\operatorname {MLC}(1)$ . Since $(f_m)|_{C^{\prime }_m}$ has the shadowing property for each $m\ge 1$ , due to Lemma 1.1, g has the shadowing property. Let us prove $h_{\mathrm {top}}(g)>0$ . Again by $\operatorname {MLC}(1)$ of $\pi '$ , we have $\hat {C}^{\prime }_n=\pi _n^{n+1}(C^{\prime }_{n+1})$ (see Lemma 2.1). Then a map $\phi \colon Y\to \pi _n^{n+1}(C^{\prime }_{n+1})$ , defined by $\phi (y)=y_n$ for all $y=(y_m)_{m\ge n}\in Y$ , gives a factor map

$$ \begin{align*} \phi\colon(Y,g)\to(\pi_n^{n+1}(C^{\prime}_{n+1}),(f_n)|_{\pi_n^{n+1}(C^{\prime}_{n+1})}). \end{align*} $$

Property (2) in the claim ensures that $\pi _n^{n+1}(C_{n+1})$ is a closed $f_n$ -invariant subset of $\pi _n^{n+1}(C^{\prime }_{n+1})$ , therefore,

$$ \begin{align*} h_{\mathrm{top}}(g)\ge h_{\mathrm{top}}((f_n)|_{\pi_n^{n+1}(C^{\prime}_{n+1})})\ge h_{\mathrm{top}}((f_n)|_{\pi_n^{n+1}(C_{n+1})})>0. \end{align*} $$

Thus, $f|_C$ has the shadowing property and satisfies $h_{\mathrm {top}}(f|_C)>0$ , completing the proof of the lemma.

Proof. We argue as in the proof of Lemma 4.1.

Step 1. Fix $D_1\in \mathcal {D}(f_1)$ with $\pi _1^2(F_2)\subset D_1$ for some $F_2\in \mathcal {D}(f_2)$ . Take $E_2\in \mathcal {D}(f_2)$ such that

(P1) $$ \begin{align} \pi_1^2(E_2)\subset D_1 \end{align} $$

and $\pi _1^2(E_2)$ is maximal among

$$ \begin{align*} \{\pi_1^2(F_2)\colon F_2\in\mathcal{D}(f_2),\pi_1^2(F_2)\subset D_1\} \end{align*} $$

with respect to the inclusion relation.

Step 2. Note that $(f_1^{m_2})|_{\pi _1^2(E_2)}$ is mixing, so $(f_1^{m_3})|_{\pi _1^2(E_2)}$ is transitive, and take a transitive point $p_1\in \pi _1^2(E_2)$ , that is, $\pi _1^2(E_2)=\omega (p_1,f_1^{m_3})$ , the $\omega $ -limit set. Since

$$ \begin{align*} p_1\in\pi_1^2(E_2)\subset\pi_1^2(X_2)=\pi_1^3(X_3), \end{align*} $$

we have $p_1=\pi _1^3(q_1)$ for some $q_1\in X_3$ . Take $D_2\in \mathcal {D}(f_2)$ with $\pi _2^3(q_1)\in D_2$ . Then choose $E_3\in \mathcal {D}(f_3)$ such that

(P2) $$ \begin{align} \pi_2^3(q_1)\in\pi_2^3(E_3)\subset D_2 \end{align} $$

and $\pi _2^3(E_3)$ is maximal among

$$ \begin{align*} \{\pi_2^3(F_3)\colon F_3\in\mathcal{D}(f_3),\pi_2^3(q_1)\in\pi_2^3(F_3)\subset D_2\} \end{align*} $$

with respect to the inclusion relation. By (P2), we have

$$ \begin{align*} p_1\in\pi_1^3(E_3)\subset\pi_1^2(D_2), \end{align*} $$

implying

$$ \begin{align*} \pi_1^2(E_2)\subset\pi_1^3(E_3)\subset\pi_1^2(D_2) \end{align*} $$

since $\pi _1^2(E_2)=\omega (p_1,f_1^{m_3})$ , and $\pi _1^3(E_3)$ is $f_1^{m_3}$ -invariant. By (P1) and $p_1\in \pi _1^2(E_2)$ , we see that $p_1\in D_1\cap \pi _1^2(D_2)$ ; therefore,

$$ \begin{align*} \pi_1^2(E_2)\subset\pi_1^3(E_3)\subset\pi_1^2(D_2)\subset D_1. \end{align*} $$

By the maximality of $\pi _1^2(E_2)$ in Step 1, we obtain

(Q1) $$ \begin{align} \pi_1^2(E_2)=\pi_1^3(E_3)=\pi_1^2(D_2). \end{align} $$

Step 3. Note that $(f_2^{m_3})|_{\pi _2^3(E_3)}$ is mixing, so $(f_2^{m_4})|_{\pi _2^3(E_3)}$ is transitive, and take a transitive point $p_2\in \pi _2^3(E_3)$ , that is, $\pi _2^3(E_3)=\omega (p_2,f_2^{m_4})$ . Since

$$ \begin{align*} p_2\in\pi_2^3(E_3)\subset\pi_2^3(X_3)=\pi_2^4(X_4), \end{align*} $$

we have $p_2=\pi _2^4(q_2)$ for some $q_2\in X_4$ . Take $D_3\in \mathcal {D}(f_3)$ with $\pi _3^4(q_2)\in D_3$ . Then choose $E_4\in \mathcal {D}(f_4)$ such that

(P3) $$ \begin{align} \pi_3^4(q_2)\in\pi_3^4(E_4)\subset D_3 \end{align} $$

and $\pi _3^4(E_4)$ is maximal among

$$ \begin{align*} \{\pi_3^4(F_4)\colon F_4\in\mathcal{D}(f_4),\pi_3^4(q_2)\in\pi_3^4(F_4)\subset D_3\} \end{align*} $$

with respect to the inclusion relation. By (P3), we have

$$ \begin{align*} p_2\in\pi_2^4(E_4)\subset\pi_2^3(D_3), \end{align*} $$

implying

$$ \begin{align*} \pi_2^3(E_3)\subset\pi_2^4(E_4)\subset\pi_2^3(D_3) \end{align*} $$

since $\pi _2^3(E_3)=\omega (p_2,f_2^{m_4})$ , and $\pi _2^4(E_4)$ is $f_2^{m_4}$ -invariant. By (P2) and $p_2\in \pi _2^3(E_3)$ , we see that $\pi _2^3(q_1)\in \pi _2^3(E_3)$ and $p_2\in D_2\cap \pi _2^3(D_3)$ ; therefore,

$$ \begin{align*} \pi_2^3(q_1)\in\pi_2^3(E_3)\subset\pi_2^4(E_4)\subset\pi_2^3(D_3)\subset D_2. \end{align*} $$

By the maximality of $\pi _2^3(E_3)$ in Step 2, we obtain

(Q2) $$ \begin{align} \pi_2^3(E_3)=\pi_2^4(E_4)=\pi_2^3(D_3). \end{align} $$

Assertions (Q1) and (Q2) yield $\pi _1^2(D_2)=\pi _1^3(D_3)$ .

Step 4. Note that $(f_3^{m_4})|_{\pi _3^4(E_4)}$ is mixing, so $(f_3^{m_5})|_{\pi _3^4(E_4)}$ is transitive, and take a transitive point $p_3\in \pi _3^4(E_4)$ , that is, $\pi _3^4(E_4)=\omega (p_3,f_3^{m_5})$ . Since

$$ \begin{align*} p_3\in\pi_3^4(E_4)\subset\pi_3^4(X_4)=\pi_3^5(X_5), \end{align*} $$

we have $p_3=\pi _3^5(q_3)$ for some $q_3\in X_5$ . Take $D_4\in \mathcal {D}(f_4)$ with $\pi _4^5(q_3)\in D_4$ . Then choose $E_5\in \mathcal {D}(f_5)$ such that

(P4) $$ \begin{align} \pi_4^5(q_3)\in\pi_4^5(E_5)\subset D_4 \end{align} $$

and $\pi _4^5(E_5)$ is maximal among

$$ \begin{align*} \{\pi_4^5(F_5)\colon F_5\in\mathcal{D}(f_5),\pi_4^5(q_3)\in\pi_4^5(F_5)\subset D_4\} \end{align*} $$

with respect to the inclusion relation. By (P4), we have

$$ \begin{align*} p_3\in\pi_3^5(E_5)\subset\pi_3^4(D_4), \end{align*} $$

implying

$$ \begin{align*} \pi_3^4(E_4)\subset\pi_3^5(E_5)\subset\pi_3^4(D_4) \end{align*} $$

since $\pi _3^4(E_4)=\omega (p_3,f_3^{m_5})$ , and $\pi _3^5(E_5)$ is $f_3^{m_5}$ -invariant. By (P3) and $p_3\in \pi _3^4(E^4)$ , we see that $\pi _3^4(q_2)\in \pi _3^4(E_4)$ and $p_3\in D_3\cap \pi _3^4(D_4)$ ; therefore,

$$ \begin{align*} \pi_3^4(q_2)\in\pi_3^4(E_4)\subset\pi_3^5(E_5)\subset\pi_3^4(D_4)\subset D_3. \end{align*} $$

By the maximality of $\pi _3^4(E_4)$ in Step 3, we obtain

(Q3) $$ \begin{align} \pi_3^4(E_4)=\pi_3^5(E_5)=\pi_3^4(D_4). \end{align} $$

Assertions (Q2) and (Q3) yield $\pi _2^3(D_3)=\pi _2^4(D_4)$ .

Continuing inductively, we obtain a sequence $D_{\ast }=(D_n)_{n\ge 1}\in \prod _{n\ge 1}\mathcal {D}(f_n)$ . For any $n\ge 1$ , $\pi _n^{n+1}(D_{n+1})\subset D_n$ and $\pi _n^{n+1}(D_{n+1})=\pi _n^{n+2}(D_{n+2})$ are established in Steps $n+1$ and $n+2$ , respectively. Thus, $D_{\ast }\in \mathcal {D}_\pi $ , and

$$ \begin{align*} \tilde{\pi}=((\pi_n^{n+1})|_{D_{n+1}}\colon D_{n+1}\to D_n)_{n\ge1} \end{align*} $$

satisfies $\operatorname {MLC}(1)$ , completing the proof.

Proof. Fix any $N>0$ and $\epsilon '>0$ . Note that $f_{N+1}\colon X_{N+1}\to X_{N+1}$ has the shadowing property, and $\mathcal {D}(f_{N+1})$ is a clopen partition of $X_{N+1}$ ; therefore, there is $\delta '>0$ such that, for any $E_{N+1}\in \mathcal {D}_{N+1}$ , every $\delta '$ -pseudo-orbit $(y_{N+1}^{(i)})_{i\ge 0}$ of $f_{N+1}$ with $y_{N+1}^{(0)}\in E_{N+1}$ is $\epsilon '$ -shadowed by some $y_{N+1}\in E_{N+1}$ .

If $\delta>0$ is small enough, then, for every $\delta $ -pseudo-orbit $\xi =(x^{(i)})_{i\ge 0}$ of f with $x^{(0)}\in D$ , $\xi _{N+1}=(x_{N+1}^{(i)})_{i\ge 0}$ is a $\delta '$ -pseudo-orbit of $f_{N+1}$ with $x_{N+1}^{(0)}\in D_{N+1}$ , which is $\epsilon '$ -shadowed by some $x_{N+1}\in D_{N+1}$ . Since

$$ \begin{align*} \pi_N^{N+1}(x_{N+1})\in\pi_N^{N+1}(D_{N+1})=\hat{D}_N \end{align*} $$

by $\operatorname {MLC}(1)$ of $\tilde {\pi }$ (see Lemma 2.1), there is $x=(x_n)_{n\ge 1}\in D=[D_{\ast }]$ such that $x_n=\pi _n^{N+1}(x_{N+1})$ for each $1\le n\le N$ . Then, for any $i\ge 0$ and $1\le n\le N$ , we have

$$ \begin{align*} \begin{aligned} d_n(f^i(x)_n,x_n^{(i)})&=d_n(f_n^i(x_n),x_n^{(i)})\\ &=d_n(f_n^i(\pi_n^{N+1}(x_{N+1})),\pi_n^{N+1}(x_{N+1}^{(i)}))\\ &=d_n(\pi_n^{N+1}(f_{N+1}^i(x_{N+1})),\pi_n^{N+1}(x_{N+1}^{(i)})) \end{aligned} \end{align*} $$

with $d_{N+1}(f_{N+1}^i(x_{N+1}),x_{N+1}^{(i)})\le \epsilon '$ . Therefore, for every $\epsilon>0$ , if N is large enough, and then $\epsilon '$ is sufficiently small, we have $d(f^i(x),x^{(i)})\le \epsilon $ for all $i\ge 0$ , that is, $\xi $ is $\epsilon $ -shadowed by $x\in D$ . Since $\xi $ is arbitrary, the lemma has been proved.

Proof. Let $D\in \mathcal {D}^{\epsilon ,\delta }(f)$ . For any $\delta $ -pseudo-orbit $\xi =(x_i)_{i\ge 0}$ of f with $x_0\in f(D)$ , take $y\in D$ with $f(y)=x_0$ and consider

$$ \begin{align*} (y,x_0,x_1,x_2,\ldots), \end{align*} $$

a $\delta $ -pseudo-orbit of f, which is $\epsilon $ -shadowed by $x\in D$ . Then $\xi $ is $\epsilon $ -shadowed by $f(x)\in f(D)$ . Since $\xi $ is arbitrary, we obtain $f(D)\in \mathcal {D}^{\epsilon ,\delta }(f)$ , proving the lemma.

Proof. Fix $D_0\in \mathcal {D}_{\mathrm {sh}}(f)\cap \mathcal {D}^{\epsilon ,\gamma }(f)$ , $x\in D_0$ , and a sequence $0<\epsilon _1>\epsilon _2>\cdots \to 0$ . Since $D_0\in \mathcal {D}_{\mathrm {sh}}(f)$ , there is a sequence $0<\delta _1>\delta _2>\cdots \to 0$ such that $D_0\in \mathcal {D}^{\epsilon _n,\delta _n}(f)$ for every $n\ge 1$ . Given any $D\in \mathcal {D}(f)$ and any $\delta $ -pseudo-orbit $\xi =(x_i)_{i\ge 0}$ of f with $x_0\in D$ , since f is chain transitive, for every $n\ge 1$ , there is a $\delta _n$ -chain $(x_i^{(n)})_{i=0}^{k_n}$ of f with $x_0^{(n)}=x$ and $x_{k_n}^{(n)}=x_0$ . Then, for each $n\ge 1$ , because $D_0\in \mathcal {D}^{\epsilon _n,\delta _n}(f)$ and $x_0^{(n)}=x\in D_0$ , we have $x_n\in D_0$ with $d(f^i(x_n),x_i^{(n)})\le \epsilon _n$ for all $0\le i\le k_n$ . Since

$$ \begin{align*} d(f^{k_n}(x_n),x_0)=d(f^{k_n}(x_n),x_{k_n}^{(n)})\le\epsilon_n, \end{align*} $$

$n\ge 1$ , $\lim _{n\to \infty }\epsilon _n=0$ , and $\delta <\gamma $ , there exists $N>0$ such that for all $n\ge N$ ,

$$ \begin{align*} \xi_n=(f^{k_n}(x_n),x_1,x_2,\ldots) \end{align*} $$

is a $\gamma $ -pseudo-orbit of f with $f^{k_n}(x_n)\in f^{k_n}(D_0)$ . For any $n\ge N$ , because $D_0\in \mathcal {D}^{\epsilon ,\gamma }(f)$ , by Lemma 5.3, we have $f^{k_n}(D_0)\in \mathcal {D}^{\epsilon ,\gamma }(f)$ , so $\xi _n$ is $\epsilon $ -shadowed by some $y_n\in f^{k_n}(D_0)$ . Taking a sequence $N\le n_1<n_2<\cdots $ such that $\lim _{j\to \infty }y_{n_j}=y$ for some $y\in X$ , we easily see that $\xi $ is $\epsilon $ -shadowed by y. Note that, for each $n\ge N$ , $\{f^{k_n}(x_n),y_n\}\subset f^{k_n}(D_0)$ and so $f^{k_n}(x_n)\sim _f y_n$ . Because $\sim _f$ is closed in $X^2$ , by

$$ \begin{align*} \lim_{j\to\infty}(f^{k_{n_j}}(x_{n_j}),y_{n_j})=(x_0,y), \end{align*} $$

we obtain $x_0\sim _f y$ and thus $y\in D$ . In other words, $\xi $ is $\epsilon $ -shadowed by $y\in D$ . Since $\xi $ and then $D\in \mathcal {D}(f)$ are arbitrary, we conclude that $\mathcal {D}(f)=\mathcal {D}^{\epsilon ,\delta }(f)$ .

Proof. Fix $\epsilon>0$ and, for any $x\in D$ , take $\delta (x,\epsilon /2)>0$ as in the assumption. Then, for each $x\in D$ , there is $0<\delta (x)<\epsilon /2$ such that, for every $\delta (x)$ -pseudo-orbit $\xi =(x_i)_{i\ge 0}$ of f with $d(x,x_0)<\delta (x)$ ,

$$ \begin{align*} \xi'=(x,x_1,x_2,x_3,\ldots) \end{align*} $$

is a $\delta (x,\epsilon /2)$ -pseudo-orbit of f, $\epsilon /2$ -shadowed by some $y\in D$ . This clearly implies that $\xi $ is $\epsilon $ -shadowed by $y\in D$ . Take a finite subset $F\subset D$ such that

$$ \begin{align*} D\subset\bigcup_{x\in F}B_{\delta(x)}(x) \end{align*} $$

where $B_{\delta (x)}(x)$ denotes the $\delta (x)$ -ball. Let $\delta =\min \{\delta (x)\colon x\in F\}$ . It follows that every $\delta $ -pseudo-orbit $(x_i)_{i\ge 0}$ of f with $x_0\in D$ is a $\delta (x)$ -pseudo-orbit of f with $x_0\in B_{\delta (x)}(x)$ for some $x\in F$ , and so $\epsilon $ -shadowed by some $y\in D$ . Since $\epsilon>0$ is arbitrary, we obtain $D\in \mathcal {D}_{\mathrm {sh}}(f)$ , proving the lemma.

Proof. Fix $D\in \mathcal {D}(f)$ . By Lemma 5.5, it suffices to show that, for any $x\in D$ and $\epsilon>0$ , there exists $\delta>0$ such that every $\delta $ -pseudo-orbit $(x_i)_{i\ge 0}$ of f with $x_0=x$ is $\epsilon $ -shadowed by some $q\in D$ , because this implies $D\in \mathcal {D}_{\mathrm {sh}}(f)$ .

Let $d_Y$ denote the metric on Y. For any $\epsilon '>0$ , Lemma 5.4 with $\mathcal {D}_{\mathrm {sh}}(g)\ne \emptyset $ implies $\mathcal {D}(g)=\mathcal {D}^{\epsilon ',\delta '}(g)$ for some $\delta '>0$ . For any $\gamma>0$ , take $0<\epsilon ''<\epsilon /2$ such that $d(x,y)\le \epsilon ''$ implies

$$ \begin{align*} \pi^{-1}(y)\subset B_\gamma(\pi^{-1}(x))=\{z\in Y\colon d_Y(z,\pi^{-1}(x))<\gamma\} \end{align*} $$

for all $y\in X$ . Since f has the shadowing property, there is $\delta>0$ such that every $\delta $ -pseudo-orbit $\xi =(x_i)_{i\ge 0}$ of f with $x_0=x$ is $\epsilon ''$ -shadowed by some $y\in X$ . Take $z\in \pi ^{-1}(y)$ and note that $d(x,y)=d(x_0,y)\le \epsilon ''$ . By the choice of $\epsilon ''$ , we obtain $w\in \pi ^{-1}(x)$ such that $d_Y(z,w)<\gamma $ . If $\gamma $ is sufficiently small, then

$$ \begin{align*} \xi'=(y_i)_{i\ge0}=(w,g(z),g^2(z),g^3(z),\ldots) \end{align*} $$

is a $\delta '$ -pseudo-orbit of g. By $\mathcal {D}(g)=\mathcal {D}^{\epsilon ',\delta '}(g)$ , $\xi '$ is $\epsilon '$ -shadowed by some $p\in Y$ with $w\sim _g p$ . Here, $w\sim _g p$ implies $\pi (w)\sim _f \pi (p)$ , so, putting $q=\pi (p)$ , we have $x\sim _f q$ , that is, $q\in D$ . Note that

$$ \begin{align*} d(q,x_0)=d(q,x)=d(\pi(p),\pi(w))=d(\pi(p),\pi(y_0)), \end{align*} $$

and

$$ \begin{align*} d(f^i(q),x_i)&\le d(f^i(q),f^i(y))+d(f^i(y),x_i)\\ &=d(f^i(\pi(p)),f^i(\pi(z)))+d(f^i(y),x_i)\\ &=d(\pi(g^i(p)),\pi(g^i(z)))+d(f^i(y),x_i)\\ &=d(\pi(g^i(p)),\pi(y_i))+d(f^i(y),x_i)\\ &\le d(\pi(g^i(p)),\pi(y_i))+\epsilon'' \\ &<d(\pi(g^i(p)),\pi(y_i))+\epsilon/2 \end{align*} $$

for each $i\ge 1$ . Since $\xi '$ is $\epsilon '$ -shadowed by p, we have $d_Y(g^i(p),y_i)\le \epsilon '$ , $i\ge 0$ , so if $\epsilon '$ is sufficiently small, then $d(f^i(q),x_i)\le \epsilon $ for all $i\ge 0$ , that is, $\xi $ is $\epsilon $ -shadowed by $q\in D$ . Since $\xi $ is arbitrary, this shows the existence of $\delta $ , and thus the lemma has been proved.

Proof. Let $\{U_n\colon n\ge 1\}$ be a countable basis for the topology of X. Then

$$ \begin{align*} D\cap T(f)=\bigcap_{n\ge1}\bigcap_{j\ge0}\bigg[D\cap\bigcup_{k\ge j}f^{-k}(U_n)\bigg], \end{align*} $$

a $G_\delta $ -subset of D. For any $n\ge 1$ , take $p_n\in U_n$ and $\epsilon _n>0$ such that

$$ \begin{align*} B_{\epsilon_n}(p_n)=\{y\in X\colon d(p_n,y)<\epsilon_n\}\subset U_n. \end{align*} $$

Let $n\ge 1$ and $j\ge 0$ . For any $x\in D$ and $0<\epsilon <\epsilon _n$ , since $D\in \mathcal {D}_{\mathrm {sh}}(f)$ , we have $D\in \mathcal {D}^{\epsilon ,\delta }(f)$ for some $\delta>0$ . Then the chain transitivity of f gives a $\delta $ -chain $(x_i)_{i=0}^k$ of f with $x_0=x$ , $x_k=p_n$ , and also $k\ge j$ . By $x\in D$ and $D\in \mathcal {D}^{\epsilon ,\delta }(f)$ , we obtain $d(f^i(y),x_i)\le \epsilon $ for all $0\le i\le k$ for some $y\in D$ . Note that $d(y,x)=d(y,x_0)\le \epsilon $ , $d(f^k(y),p_n)=d(f^k(y),x_k)\le \epsilon <\epsilon _n$ , and so $f^k(y)\in U_n$ , implying

$$ \begin{align*} y\in D\cap\bigcup_{k\ge j}f^{-k}(U_n). \end{align*} $$

Since $x\in D$ and $0<\epsilon <\epsilon _n$ are arbitrary, this shows that

$$ \begin{align*} D\cap\bigcup_{k\ge j}f^{-k}(U_n) \end{align*} $$

is dense in D. Since $n\ge 1$ and $j\ge 0$ are arbitrary, we conclude that $D\cap T(f)$ is a dense $G_\delta $ -subset of D, completing the proof.

Proof. Given any $\epsilon>0$ , take $\delta>0$ so small that $D\in \mathcal {D}^{\epsilon ,\delta }(f)$ . For this $\delta $ , choose $N>0$ as in property (3) of $\sim _{f,\delta }$ (see §2.2.3). Note that $y,z\in D$ implies $y\sim _f z$ and so $y\sim _{f,\delta }z$ . Since $y\sim _{f,\delta }f^{mN}(y)$ , we have $z\sim _{f,\delta }f^{mN}(y)$ . Then the choice of N gives a $\delta $ -chain $\alpha =(y_i)_{i=0}^{mN}$ of f with $y_0=z$ and $y_{mN}=f^{mN}(y)$ . Let

$$ \begin{align*} \beta=(f^{mN}(y),f^{mN+1}(y),\ldots) \end{align*} $$

and $\xi =\alpha \beta =(x_i)_{i\ge 0}$ . Then $\xi $ is a $\delta $ -pseudo-orbit of f with $x_0=z\in D$ , so is $\epsilon $ -shadowed by some $w\in D$ . Note that $d(z,w)=d(x_0,w)\le \epsilon $ . Also, we have

$$ \begin{align*} d(f^i(y),f^i(w))=d(x_i,f^i(w))\le\epsilon \end{align*} $$

for every $i\ge mN$ , so $\limsup _{k\to \infty }d(f^k(y),f^k(w))\le \epsilon $ . This completes the proof.