[proofplan] The proof identifies points in the maximal invariant set with their bi-infinite strip itineraries. The disjointness and relative clopen hypotheses for the strip pieces in $\Lambda$ make the itinerary map well-defined and locally constant on finite symbolic coordinates. The explicit cylinder hypotheses imply that every finite symbolic prescription determines a nonempty compact cylinder rectangle and that nested cylinder rectangles shrink to one point, which gives bijectivity. The conjugacy relation follows by shifting time indices, and the two continuity statements follow from finite cylinder conditions and the shrinking-diameter conclusion. [/proofplan] [step:Show that every point in $\Lambda$ has a unique bi-infinite itinerary] Let $x \in \Lambda$ and let $n \in \mathbb{Z}$. Since $\Lambda$ is defined by \begin{align*} \Lambda = \bigcap_{m \in \mathbb{Z}} F^{-m}(R), \end{align*} we have $F^n(x) \in R$ and $F^{n+1}(x) \in R$. Hence $F^n(x) \in R \cap F^{-1}(R)$. By the strip decomposition, \begin{align*} R \cap F^{-1}(R) = H_0 \cup H_1, \end{align*} so $F^n(x) \in H_0 \cup H_1$. The condition $H_0 \cap H_1 \cap \Lambda = \varnothing$ implies that no point of $\Lambda$ belongs to both strips. Since $F^n(x) \in \Lambda$ by invariance of $\Lambda$, there is a unique symbol $i \in \{0,1\}$ such that $F^n(x) \in H_i$. Thus the rule \begin{align*} \pi(x)_n = i \quad \text{if and only if} \quad F^n(x) \in H_i \end{align*} defines a map $\pi:\Lambda \to \Sigma_2$. [/step] [step:Use cylinder rectangles to realize and separate symbolic sequences] For integers $N \geq 0$ and a finite word \begin{align*} w = (w_{-N},\dots,w_N) \in \{0,1\}^{\{-N,\dots,N\}}, \end{align*} define the finite cylinder rectangle in phase space by \begin{align*} C_N(w) := \{x \in R : F^k(x) \in H_{w_k} \text{ for every } k \in \{-N,\dots,N\}\}. \end{align*} This definition is meaningful for negative $k$ because the repaired statement assumes that $F:U\to U$ is a $C^1$ diffeomorphism, so every iterate $F^k:U\to U$ is defined for $k\in\mathbb{Z}$. Let $\mathcal{B}_{\mathrm{nb}}(\mathbb{R}^2)$ denote the set of all nonempty bounded subsets of $\mathbb{R}^2$. The diameter map $\operatorname{diam}:\mathcal{B}_{\mathrm{nb}}(\mathbb{R}^2)\to[0,\infty)$ is defined by \begin{align*} \operatorname{diam}(A) := \sup\{|x-y|:x,y\in A\}. \end{align*} The cylinder conclusions included in the theorem statement give the facts needed below. First, for every $N \geq 0$ and every word $w \in \{0,1\}^{\{-N,\dots,N\}}$, the set $C_N(w)$ is nonempty and compact. Second, if $s=(s_n)_{n \in \mathbb{Z}} \in \Sigma_2$ and $w_N=(s_{-N},\dots,s_N)$, then the compact sets $C_N(w_N)$ are nested and their diameters tend to zero: \begin{align*} C_{N+1}(w_{N+1}) \subset C_N(w_N) \end{align*} and \begin{align*} \lim_{N \to \infty} \operatorname{diam}(C_N(w_N)) = 0. \end{align*} Because $R$ is compact and the sets $C_N(w_N)$ are nonempty nested compact subsets of $R$, the finite intersection property for compact spaces implies that their intersection is nonempty. Since their diameters tend to zero, their intersection contains at most one point. Therefore there is a unique point $x_s \in R$ such that \begin{align*} \{x_s\} = \bigcap_{N=0}^{\infty} C_N(w_N). \end{align*} For every $k \in \mathbb{Z}$, choosing $N \geq |k|$ gives $F^k(x_s) \in H_{s_k} \subset R$. Hence $x_s \in \Lambda$, and $\pi(x_s)=s$. [guided] Fix a bi-infinite symbolic sequence $s=(s_n)_{n \in \mathbb{Z}} \in \Sigma_2$. The goal is to construct the unique point whose orbit visits the strip $H_{s_n}$ at time $n$. Since we cannot impose infinitely many strip conditions at once directly, we first impose only finitely many of them. For each integer $N \geq 0$, let \begin{align*} w_N = (s_{-N},\dots,s_N) \in \{0,1\}^{\{-N,\dots,N\}}. \end{align*} Define \begin{align*} C_N(w_N) := \{x \in R : F^k(x) \in H_{s_k} \text{ for every } k \in \{-N,\dots,N\}\}. \end{align*} This set consists of all points whose orbit follows the prescribed symbols from time $-N$ through time $N$. We now use the cylinder conclusions included explicitly in the theorem statement. They say that every finite word is admissible, the corresponding cylinder is nonempty and compact, the cylinders for a fixed bi-infinite sequence are nested, and their diameters tend to zero. These are the exact structural inputs needed to turn symbolic data into a unique orbit point. The theorem gives two facts for the sets $C_N(w_N)$. First, each $C_N(w_N)$ is a nonempty compact subset of $R$. Second, the sequence is nested and shrinking: \begin{align*} C_{N+1}(w_{N+1}) \subset C_N(w_N) \end{align*} and \begin{align*} \lim_{N \to \infty} \operatorname{diam}(C_N(w_N)) = 0. \end{align*} The nesting holds because imposing the symbols from time $-(N+1)$ through time $N+1$ includes, in particular, imposing the symbols from time $-N$ through time $N$. Since $R$ is compact and the sets $C_N(w_N)$ are nonempty nested compact subsets of $R$, the finite intersection property for compact spaces gives \begin{align*} \bigcap_{N=0}^{\infty} C_N(w_N) \neq \varnothing. \end{align*} The shrinking diameter conclusion gives uniqueness. Indeed, if $x$ and $y$ both belonged to the intersection, then for every $N$ both points would lie in $C_N(w_N)$, so \begin{align*} |x-y| \leq \operatorname{diam}(C_N(w_N)). \end{align*} Taking $N \to \infty$ gives $|x-y|=0$, hence $x=y$. Denote the unique point in the intersection by $x_s$. For any fixed $k \in \mathbb{Z}$, choose $N \geq |k|$. Since $x_s \in C_N(w_N)$, the definition of $C_N(w_N)$ gives $F^k(x_s) \in H_{s_k} \subset R$. This is true for every $k \in \mathbb{Z}$, so \begin{align*} x_s \in \bigcap_{k \in \mathbb{Z}} F^{-k}(R) = \Lambda. \end{align*} By construction, the itinerary of $x_s$ is exactly $s$, so $\pi(x_s)=s$. [/guided] [/step] [step:Conclude that the itinerary map is bijective] The construction in the previous step proves surjectivity: for every $s \in \Sigma_2$, the point $x_s \in \Lambda$ satisfies $\pi(x_s)=s$. To prove injectivity, let $x,y \in \Lambda$ and suppose $\pi(x)=\pi(y)=s$. For each $N \geq 0$, both $x$ and $y$ belong to the cylinder rectangle $C_N(w_N)$ associated to $w_N=(s_{-N},\dots,s_N)$. Hence \begin{align*} |x-y| \leq \operatorname{diam}(C_N(w_N)). \end{align*} The shrinking cylinder conclusion assumed in the theorem statement gives \begin{align*} \lim_{N \to \infty} \operatorname{diam}(C_N(w_N)) = 0, \end{align*} so $|x-y|=0$. Therefore $x=y$, and $\pi$ is injective. Thus $\pi:\Lambda \to \Sigma_2$ is bijective. [/step] [step:Verify that the itinerary map intertwines $F$ with the shift] Let $x \in \Lambda$ and let $n \in \mathbb{Z}$. By definition of $\pi$, \begin{align*} (\pi(F(x)))_n = i \quad \text{if and only if} \quad F^n(F(x)) \in H_i. \end{align*} Since $F^n(F(x))=F^{n+1}(x)$, this is equivalent to \begin{align*} \pi(x)_{n+1}=i. \end{align*} By definition of the shift map $\sigma:\Sigma_2 \to \Sigma_2$, \begin{align*} (\sigma(\pi(x)))_n = \pi(x)_{n+1}. \end{align*} Therefore \begin{align*} (\pi(F(x)))_n = (\sigma(\pi(x)))_n \end{align*} for every $n \in \mathbb{Z}$. Hence $\pi(F(x))=\sigma(\pi(x))$, and since $x \in \Lambda$ was arbitrary, \begin{align*} \pi \circ F = \sigma \circ \pi. \end{align*} [/step] [step:Prove that $\pi$ and its inverse are continuous] We first prove continuity of $\pi$. Let $x \in \Lambda$ and let $M \geq 0$. A basic neighbourhood of $\pi(x)$ in $\Sigma_2$ is determined by requiring agreement with $\pi(x)$ on the finite set $\{-M,\dots,M\}$. For each $k \in \{-M,\dots,M\}$, the point $F^k(x)$ lies in exactly one of the two sets $H_0\cap\Lambda$ and $H_1\cap\Lambda$. By the repaired statement, both $H_0\cap\Lambda$ and $H_1\cap\Lambda$ are relatively open in $\Lambda$. Since $F^k|_{\Lambda}:\Lambda\to\Lambda$ is continuous for each $k\in\mathbb{Z}$, the set of points $y\in\Lambda$ for which $F^k(y)$ lies in the same strip as $F^k(x)$ is a neighbourhood of $x$ in $\Lambda$. Intersecting these finitely many neighbourhoods over $k\in\{-M,\dots,M\}$ gives a neighbourhood $O_x \subset \Lambda$ of $x$ such that for every $y \in O_x$ and every $k \in \{-M,\dots,M\}$, the points $F^k(y)$ and $F^k(x)$ lie in the same strip. Hence \begin{align*} \pi(y)_k=\pi(x)_k \end{align*} for every $k \in \{-M,\dots,M\}$, proving continuity of $\pi$. It remains to prove continuity of $\pi^{-1}:\Sigma_2 \to \Lambda$. Let $s \in \Sigma_2$ and let $\varepsilon>0$. By the shrinking-diameter conclusion, there exists $N \geq 0$ such that the cylinder rectangle $C_N(w_N)$ for $w_N=(s_{-N},\dots,s_N)$ satisfies \begin{align*} \operatorname{diam}(C_N(w_N)) < \varepsilon. \end{align*} If $t \in \Sigma_2$ agrees with $s$ on $\{-N,\dots,N\}$, then both $\pi^{-1}(s)$ and $\pi^{-1}(t)$ belong to the same set $C_N(w_N)$. Therefore \begin{align*} |\pi^{-1}(s)-\pi^{-1}(t)| \leq \operatorname{diam}(C_N(w_N)) < \varepsilon. \end{align*} The set of all such $t$ is a basic neighbourhood of $s$ in the [product topology](/page/Product%20Topology) on $\Sigma_2$, so $\pi^{-1}$ is continuous at $s$. Since $s$ was arbitrary, $\pi^{-1}$ is continuous on $\Sigma_2$. Thus $\pi$ is a continuous bijection with continuous inverse, so $\pi:\Lambda \to \Sigma_2$ is a homeomorphism. Together with the identity $\pi \circ F=\sigma \circ \pi$, this proves that $F|_{\Lambda}$ is topologically conjugate to the full two-shift. [/step]