[proofplan] The proof is a symbolic coding argument followed by the standard passage from a return map to a suspension flow. For each bi-infinite binary sequence, the finite central cylinder sets form a nested family of nonempty compact sets whose diameters shrink uniformly to zero, so their intersection contains exactly one point. This gives a coding map between the invariant set $K$ and the full two-shift; disjointness of $R_0$ and $R_1$ gives uniqueness of itineraries, while compactness and uniform shrinking give continuity. Finally, the continuous positive roof function turns the base conjugacy into a suspension conjugacy, and the assumed injectivity of the evaluation map identifies the suspension with the invariant flow set $\Lambda$. [/proofplan] [step:Construct one point for each bi-infinite symbolic itinerary] Fix a metric $d_R:R\times R\to[0,\infty)$ inducing the topology of the [compact space](/page/Compact%20Space) $R$. Let \begin{align*} \Omega:=\{0,1\}^{\mathbb{Z}} \end{align*} with the [product topology](/page/Product%20Topology), and let $\sigma:\Omega\to\Omega$ be the left shift, defined by $(\sigma a)_j=a_{j+1}$ for $a=(a_j)_{j\in\mathbb{Z}}\in\Omega$. For $a=(a_j)_{j\in\mathbb{Z}}\in\Omega$ and $N\in\mathbb{N}$, define \begin{align*} C_N(a):=C(a_{-N},\dots,a_N). \end{align*} If $N_1\leq N_2$, then $C_{N_2}(a)\subset C_{N_1}(a)$, because the conditions defining $C_{N_2}(a)$ include all the conditions defining $C_{N_1}(a)$. Each $C_N(a)$ is nonempty and compact by hypothesis. Since $R$ is compact, the nested compact intersection principle gives \begin{align*} \bigcap_{N=1}^{\infty} C_N(a)\neq\varnothing. \end{align*} The uniform diameter hypothesis gives $\operatorname{diam} C_N(a)\to 0$. Hence this intersection contains exactly one point: if $x$ and $y$ both lie in the intersection, then $d_R(x,y)\leq \operatorname{diam} C_N(a)$ for every $N$, and therefore $d_R(x,y)=0$. Define the map $q:\Omega\to R$ by declaring $q(a)$ to be the unique point in $\bigcap_{N=1}^{\infty}C_N(a)$. [guided] The coding construction starts from the finite cylinders because the hypotheses guarantee compactness and nonemptiness only for finite words. Fix a metric $d_R:R\times R\to[0,\infty)$ inducing the topology on $R$. This is legitimate because the statement assumes the shrinking property for any such metric. Let $\Omega:=\{0,1\}^{\mathbb{Z}}$ be the space of all bi-infinite binary sequences, and define the left shift $\sigma:\Omega\to\Omega$ by $(\sigma a)_j=a_{j+1}$. For a sequence $a=(a_j)_{j\in\mathbb{Z}}$ and an integer $N\in\mathbb{N}$, set \begin{align*} C_N(a):=C(a_{-N},\dots,a_N). \end{align*} The point of $C_N(a)$ is that it records the requested itinerary only from time $-N$ to time $N$. If $N_1\leq N_2$, then every point whose itinerary matches $a$ from $-N_2$ to $N_2$ also matches $a$ from $-N_1$ to $N_1$. Therefore \begin{align*} C_{N_2}(a)\subset C_{N_1}(a). \end{align*} Thus the sets $C_N(a)$ form a nested sequence of nonempty compact subsets of the compact space $R$. The nested compact intersection principle implies that their intersection is nonempty: \begin{align*} \bigcap_{N=1}^{\infty}C_N(a)\neq\varnothing. \end{align*} We now use the shrinking hypothesis to prove uniqueness. Suppose $x,y\in\bigcap_{N=1}^{\infty}C_N(a)$. Then $x,y\in C_N(a)$ for every $N$, so \begin{align*} d_R(x,y)\leq \operatorname{diam} C_N(a) \end{align*} for every $N$. The diameters tend to $0$ uniformly, hence in particular they tend to $0$ along this fixed sequence $a$. Therefore $d_R(x,y)=0$, so $x=y$. This proves that every symbolic sequence determines exactly one point of $R$. Define $q:\Omega\to R$ by taking $q(a)$ to be that unique point. [/guided] [/step] [step:Identify the image of the coding map with the invariant set] We first show $q(\Omega)\subset K$. Let $a=(a_j)_{j\in\mathbb{Z}}\in\Omega$ and set $x:=q(a)$. Fix $n\in\mathbb{Z}$. Choose $N\in\mathbb{N}$ with $N\geq |n|$. Since $x\in C_N(a)$, the iterate $F^n(x)$ is defined and lies in $R_{a_n}\subset R_0\cup R_1$. Since this holds for every $n\in\mathbb{Z}$, $x\in K$. Conversely, let $x\in K$. Since $R_0$ and $R_1$ are disjoint and $F^j(x)\in R_0\cup R_1$ for every $j\in\mathbb{Z}$, there is a unique sequence $a(x)=(a_j(x))_{j\in\mathbb{Z}}\in\Omega$ such that \begin{align*} F^j(x)\in R_{a_j(x)} \end{align*} for every $j\in\mathbb{Z}$. Then $x\in C_N(a(x))$ for every $N\in\mathbb{N}$, so by the uniqueness of the intersection defining $q(a(x))$ we have $q(a(x))=x$. Hence \begin{align*} q(\Omega)=K. \end{align*} In particular, $K$ is nonempty. [/step] [step:Prove that the itinerary map is a conjugacy with the full shift] Define $h:K\to\Omega$ by \begin{align*} h(x):=(a_j(x))_{j\in\mathbb{Z}}, \end{align*} where $a_j(x)\in\{0,1\}$ is the unique symbol satisfying $F^j(x)\in R_{a_j(x)}$. The preceding step shows that $q:\Omega\to K$ and $h:K\to\Omega$ are inverse maps, so $h$ is bijective. For $x\in K$ and $j\in\mathbb{Z}$, the iterate $F^j(F(x))$ is defined and equals $F^{j+1}(x)$. Therefore \begin{align*} h(F(x))_j=h(x)_{j+1}=(\sigma h(x))_j. \end{align*} Thus $h\circ F=\sigma\circ h$. This identity also shows $F(K)\subset K$. Since $\sigma$ is surjective and $h$ is bijective, for each $y\in K$ there is $x:=h^{-1}(\sigma^{-1}h(y))\in K$ with $F(x)=y$, so $F(K)=K$. It remains to prove continuity. Let $a_r$ be a net in $\Omega$ converging to $a\in\Omega$. Fix $\varepsilon>0$. Choose $N\in\mathbb{N}$ such that every cylinder $C(b_{-N},\dots,b_N)$ has $d_R$-diameter less than $\varepsilon$. Since $a_r\to a$ in the product topology, eventually $(a_r)_j=a_j$ for all $-N\leq j\leq N$. For all such $r$, both $q(a_r)$ and $q(a)$ lie in $C_N(a)$, hence \begin{align*} d_R(q(a^{(r)}),q(a))<\varepsilon. \end{align*} Thus $q:\Omega\to K$ is continuous. Since $\Omega$ is compact by the product [compactness theorem](/theorems/2748) and $R$ is Hausdorff, $K=q(\Omega)$ is compact and $q:\Omega\to K$ is a homeomorphism. Therefore $h=q^{-1}$ is a homeomorphism. [guided] The itinerary map $h$ records which rectangle contains each iterate. For $x\in K$, every iterate $F^j(x)$ is defined and lies in $R_0\cup R_1$. Because $R_0$ and $R_1$ are disjoint, there is exactly one symbol $a_j(x)\in\{0,1\}$ such that \begin{align*} F^j(x)\in R_{a_j(x)}. \end{align*} Define $h:K\to\Omega$ by $h(x)=(a_j(x))_{j\in\mathbb{Z}}$. The previous step proves that every symbolic sequence $a\in\Omega$ has a unique realized point $q(a)\in K$, and that every $x\in K$ is realized by its own itinerary. Therefore $h$ and $q$ are inverse maps. Now check the dynamics. If $x\in K$, then for every $j\in\mathbb{Z}$ the point $F^j(F(x))$ is defined and equals $F^{j+1}(x)$. Hence the $j$-th symbol of the itinerary of $F(x)$ is the $(j+1)$-st symbol of the itinerary of $x$: \begin{align*} h(F(x))_j=h(x)_{j+1}. \end{align*} The left shift $\sigma$ is defined by $(\sigma b)_j=b_{j+1}$, so this is exactly \begin{align*} h(F(x))_j=(\sigma h(x))_j. \end{align*} Since this holds for every coordinate $j$, we have $h\circ F=\sigma\circ h$. This also proves invariance. The inclusion $F(K)\subset K$ follows because $h(F(x))$ is defined for every $x\in K$. For the reverse inclusion, take $y\in K$. Since $\sigma$ is onto, choose $b\in\Omega$ with $\sigma b=h(y)$. Let $x:=h^{-1}(b)$. Then \begin{align*} h(F(x))=\sigma h(x)=\sigma b=h(y). \end{align*} The map $h$ is injective, so $F(x)=y$. Hence $K\subset F(K)$, and therefore $F(K)=K$. Finally we prove continuity using the uniform shrinking of cylinders. Let $a^{(r)}$ be a net in $\Omega$ converging to $a$. Nets are used because this argument works in every compact topological setting; in the metrizable product space $\Omega$, sequences would also suffice. Fix $\varepsilon>0$. By uniform shrinking, choose $N\in\mathbb{N}$ such that every cylinder of length $2N+1$ has $d_R$-diameter less than $\varepsilon$. Convergence in the product topology means that eventually the central block agrees: \begin{align*} a^{(r)}_j=a_j \quad \text{for all } -N\leq j\leq N. \end{align*} For such $r$, the points $q(a^{(r)})$ and $q(a)$ both belong to the same compact cylinder $C_N(a)$. Therefore \begin{align*} d_R(q(a^{(r)}),q(a))<\varepsilon. \end{align*} Thus $q$ is continuous. The space $\Omega=\{0,1\}^{\mathbb{Z}}$ is compact by the product compactness theorem, and $R$ is Hausdorff because it is metrizable. Hence the continuous bijection $q:\Omega\to K$ is a homeomorphism onto its image, and $K=q(\Omega)$ is compact. Therefore $h=q^{-1}$ is also continuous, so $h$ is a homeomorphism. [/guided] [/step] [step:Prove continuity and positivity of the roof function on the invariant set] For each integer $j$ with $0\leq j\leq m-1$, define \begin{align*} \rho_j:K\to(0,\infty) \end{align*} by $\rho_j(x):=\tau(P^j(x))$. The map $P^j$ is continuous on $K$ because it is an iterate of the homeomorphism $P$ on the subset where the iterate is defined, and by hypothesis $\tau$ is continuous at every point of $P^j(R_0\cup R_1)$. Since $K\subset R_0\cup R_1$, each $\rho_j$ is continuous on $K$. The roof function restricted to $K$ is the finite sum \begin{align*} \tau_m|_K=\sum_{j=0}^{m-1}\rho_j. \end{align*} Therefore $\tau_m|_K$ is continuous. Since each value of $\tau$ lies in $(0,\infty)$, every summand $\rho_j$ is positive, and hence $\tau_m(x)>0$ for every $x\in K$. [/step] [step:Realize the invariant set as the compact suspension image] Define \begin{align*} S:=\{(x,s)\in K\times\mathbb{R}:0\leq s\leq \tau_m(x)\}. \end{align*} Since $K$ is compact and $\tau_m|_K$ is continuous, the set $S$ is compact. The map $\widetilde E:S\to M$ defined by \begin{align*} \widetilde E(x,s):=\Phi_s(x) \end{align*} is continuous because $\Phi$ is continuous. By definition, \begin{align*} \widetilde E(S)=\Lambda. \end{align*} Thus $\Lambda$ is compact and nonempty. The assumed containment of all orbit segments from $R_0\cup R_1$ gives $\Lambda\subset U$ because $K\subset R_0\cup R_1$. Let $\pi:K\times\mathbb{R}\to K_{\tau_m}$ be the quotient map. The map $E:K_{\tau_m}\to M$ is assumed well-defined and satisfies $E(\pi(x,t))=\Phi_t(x)$. Its image is $\Lambda$, because every class has a representative $(x,s)$ with $x\in K$ and $0\leq s\leq\tau_m(x)$ by repeatedly using the relation $(x,t+\tau_m(x))\sim(F(x),t)$ and the invariance $F(K)=K$. Hence \begin{align*} E(K_{\tau_m})=\Lambda. \end{align*} The suspension space $K_{\tau_m}$ is compact, since it is the quotient of the compact fundamental domain $S$ after identifying the endpoint pairs $(x,\tau_m(x))$ and $(F(x),0)$. The map $E:K_{\tau_m}\to\Lambda$ is a continuous bijection by the assumed injectivity and the preceding image identity. Since $K_{\tau_m}$ is compact and $M$ is Hausdorff, $E$ is a homeomorphism onto $\Lambda$. [/step] [step:Conjugate the suspension flow to the restricted flow on $\Lambda$] Let $\Psi:\mathbb{R}\times K_{\tau_m}\to K_{\tau_m}$ denote the suspension flow, defined by \begin{align*} \Psi_s([(x,t)]):=[(x,t+s)]. \end{align*} For $s\in\mathbb{R}$ and $[(x,t)]\in K_{\tau_m}$, the flow property of $\Phi$ gives \begin{align*} E(\Psi_s([(x,t)]))=E([(x,t+s)])=\Phi_{t+s}(x)=\Phi_s(\Phi_t(x))=\Phi_s(E([(x,t)])). \end{align*} Thus $E$ conjugates the suspension flow to the restriction of $\Phi$ on $\Lambda$. This identity also implies that $\Lambda$ is $\Phi$-invariant. Finally, since $h:K\to\Omega$ is a homeomorphism satisfying $h\circ F=\sigma\circ h$, the map induced by $(x,t)\mapsto(h(x),t)$ is a homeomorphism from the suspension over $F|_K$ with roof $\tau_m|_K$ to the suspension over $(\Omega,\sigma)$ with roof $\tau_m\circ h^{-1}$. Composing this suspension homeomorphism with $E$ gives the stated topological conjugacy between the restricted flow on $\Lambda$ and the suspension flow over the full two-shift with roof function $\tau_m\circ h^{-1}$. This completes the proof. [/step]