[proofplan] The symbolic coding conclusion identifies the horseshoe dynamics with the full two-sided shift on two symbols. We prove the four dynamical properties directly for the full shift: transitivity by inserting two prescribed finite blocks into one bi-infinite sequence, density of periodic points by periodically repeating the finite word defining a cylinder, sensitivity by forcing a future disagreement at the zero coordinate, and entropy by counting binary words. The conjugacy then transfers transitivity, periodic-point density, sensitivity, and topological entropy from the full shift back to $F|_{\Lambda}$. The branch-crossing-only clause is the same argument restricted to the conjugate invariant subsystem, without making claims about any larger maximal invariant set. [/proofplan]

[step:Pass from the horseshoe to the full two-shift by symbolic coding] Let \begin{align*} \Sigma_2 := \{0,1\}^{\mathbb{Z}} \end{align*} with the [product topology](/page/Product%20Topology), and define the full two-sided shift \begin{align*} \sigma: \Sigma_2 \to \Sigma_2 \end{align*} by \begin{align*} (\sigma a)_k := a_{k+1} \end{align*} for every sequence $a = (a_k)_{k \in \mathbb{Z}} \in \Sigma_2$ and every index $k \in \mathbb{Z}$. By the explicitly assumed horseshoe coding data, there is a homeomorphism \begin{align*} h: \Lambda \to \Sigma_2 \end{align*} such that $h \circ F|_{\Lambda} = \sigma \circ h$. Thus $F|_{\Lambda}$ and $\sigma$ are topologically conjugate. It remains to prove the stated properties for $\sigma$ and then transfer them through $h$. [/step]

[step:Construct a point whose orbit visits any two prescribed cylinders]A cylinder in $\Sigma_2$ is a set determined by finitely many coordinate conditions. Let $U,V \subset \Sigma_2$ be nonempty open sets. Choose nonempty cylinders $D_U \subset U$ and $D_V \subset V$. Since $D_U$ and $D_V$ are nonempty, choose points $c \in D_U$ and $d \in D_V$. Let $I_U \subset \mathbb{Z}$ and $I_V \subset \mathbb{Z}$ be finite coordinate sets defining $D_U$ and $D_V$, respectively. Choose integers $r \le s$ and $p \le q$ such that $I_U \subset \{r,\dots,s\}$ and $I_V \subset \{p,\dots,q\}$. Define the interval cylinders \begin{align*} C_U := \{a \in \Sigma_2 : a_i = c_i \text{ for every } i \in \{r,\dots,s\}\} \end{align*} and \begin{align*} C_V := \{a \in \Sigma_2 : a_j = d_j \text{ for every } j \in \{p,\dots,q\}\}. \end{align*} Then $C_U$ and $C_V$ are nonempty, $C_U \subset D_U \subset U$, and $C_V \subset D_V \subset V$. Choose an integer $n \ge 0$ so large that the shifted coordinate interval $\{p+n,\dots,q+n\}$ is disjoint from $\{r,\dots,s\}$. Define a sequence $a = (a_k)_{k \in \mathbb{Z}} \in \Sigma_2$ by setting $a_i := c_i$ for $i \in \{r,\dots,s\}$, setting $a_{j+n} := d_j$ for $j \in \{p,\dots,q\}$, and assigning arbitrary values in $\{0,1\}$ to all remaining coordinates. The disjointness of the two coordinate intervals makes this definition consistent. Then $a \in C_U \subset U$. Also, for every $j \in \{p,\dots,q\}$, \begin{align*} (\sigma^n a)_j = a_{j+n} = d_j. \end{align*} Thus $\sigma^n a \in C_V \subset V$. Hence $\sigma^n(U) \cap V \ne \varnothing$, so $\sigma$ is topologically transitive.[/step]

[guided]We prove transitivity using only cylinder sets for the product topology, but we must be careful about what a cylinder means. A general cylinder may constrain an arbitrary finite set of coordinates, not necessarily a contiguous interval. The definition of topological transitivity asks: given two nonempty open sets $U$ and $V$, can we find one orbit that starts in $U$ and later enters $V$? Since cylinders form a basis, choose nonempty cylinders $D_U \subset U$ and $D_V \subset V$. Let $I_U \subset \mathbb{Z}$ be the finite coordinate set defining $D_U$, and let $I_V \subset \mathbb{Z}$ be the finite coordinate set defining $D_V$. Choose points $c \in D_U$ and $d \in D_V$. We now pass to smaller interval cylinders, because interval cylinders are convenient for shifting without conflicts. Choose integers $r \le s$ and $p \le q$ such that \begin{align*} I_U \subset \{r,\dots,s\} \end{align*} and \begin{align*} I_V \subset \{p,\dots,q\}. \end{align*} Define \begin{align*} C_U := \{a \in \Sigma_2 : a_i = c_i \text{ for every } i \in \{r,\dots,s\}\} \end{align*} and \begin{align*} C_V := \{a \in \Sigma_2 : a_j = d_j \text{ for every } j \in \{p,\dots,q\}\}. \end{align*} These are nonempty because they contain $c$ and $d$, respectively. Also $C_U \subset D_U \subset U$ and $C_V \subset D_V \subset V$, because the interval-cylinder conditions include all the original finite coordinate conditions. The word defining $C_U$ imposes finitely many constraints on the coordinates of $a$. The condition $\sigma^n a \in C_V$ imposes finitely many constraints on a shifted set of coordinates of $a$, namely the coordinates $j+n$ with $j \in \{p,\dots,q\}$, because $(\sigma^n a)_j = a_{j+n}$. Now choose $n \ge 0$ so large that the finite interval $\{p+n,\dots,q+n\}$ is disjoint from $\{r,\dots,s\}$. After shifting far enough, the coordinates needed to force the future visit to $V$ do not conflict with the coordinates needed to start inside $U$. Define $a \in \Sigma_2$ by \begin{align*} a_i := c_i \text{ for } i \in \{r,\dots,s\} \end{align*} and \begin{align*} a_{j+n} := d_j \text{ for } j \in \{p,\dots,q\}. \end{align*} All other coordinates are chosen arbitrarily in $\{0,1\}$. Because the two finite coordinate sets are disjoint, these assignments are compatible. The first assignment gives $a \in C_U \subset U$. The second assignment gives, for every $j \in \{p,\dots,q\}$, \begin{align*} (\sigma^n a)_j = a_{j+n} = d_j. \end{align*} Thus $\sigma^n a \in C_V \subset V$. Therefore $\sigma^n(U) \cap V \ne \varnothing$, which is exactly topological transitivity.[/guided]

[step:Approximate every cylinder by a periodic sequence] Let $O \subset \Sigma_2$ be a nonempty [open set](/page/Open%20Set). Choose a nonempty cylinder $D \subset O$, let $I \subset \mathbb{Z}$ be the finite coordinate set defining $D$, and choose a point $c \in D$. Choose integers $r \le s$ such that $I \subset \{r,\dots,s\}$, and define the nonempty interval cylinder \begin{align*} C := \{a \in \Sigma_2 : a_i = c_i \text{ for every } i \in \{r,\dots,s\}\}. \end{align*} Then $C \subset D \subset O$. Let $L := s-r+1$. Define a sequence $z = (z_k)_{k \in \mathbb{Z}} \in \Sigma_2$ by repeating the block $(c_r,\dots,c_s)$ with period $L$, aligned so that $z_i = c_i$ for every $i \in \{r,\dots,s\}$. Then $z \in C \subset O$ and $\sigma^L z = z$. Hence every nonempty open set contains a periodic point, so the periodic points of $\sigma$ are dense in $\Sigma_2$. [/step]

[step:Force a future separation from any neighbourhood] Equip $\Sigma_2$ with the compatible metric. Define the map \begin{align*} d_{\Sigma}: \Sigma_2 \times \Sigma_2 \to \mathbb{R} \end{align*} by \begin{align*} (a,b) \mapsto \sum_{k \in \mathbb{Z}} 2^{-|k|}\, |a_k-b_k| \end{align*} for $a,b \in \Sigma_2$. Let $a \in \Sigma_2$ and let $U \subset \Sigma_2$ be an open neighbourhood of $a$. Choose a cylinder $C \subset U$ containing $a$, determined by the coordinates in a finite interval $\{-N,\dots,N\}$ for some $N \in \mathbb{N}$. Define $b = (b_k)_{k \in \mathbb{Z}} \in \Sigma_2$ by requiring $b_k = a_k$ for every $k \in \{-N,\dots,N\}$, requiring $b_{N+1} = 1-a_{N+1}$, and assigning arbitrary values in $\{0,1\}$ to all other coordinates. Then $b \in C \subset U$. At time $N+1$, \begin{align*} (\sigma^{N+1}a)_0 = a_{N+1} \end{align*} and \begin{align*} (\sigma^{N+1}b)_0 = b_{N+1} = 1-a_{N+1}. \end{align*} Therefore \begin{align*} d_{\Sigma}(\sigma^{N+1}a,\sigma^{N+1}b) \ge |(\sigma^{N+1}a)_0-(\sigma^{N+1}b)_0| = 1. \end{align*} Thus $\sigma$ has sensitive dependence on initial conditions, with sensitivity constant $\delta := 1/2$. [/step]

[step:Compute the entropy of the full two-shift by counting words] For each $n \in \mathbb{N}$, let $\mathcal{L}_n(\Sigma_2)$ denote the set of length-$n$ admissible words in the subshift $\Sigma_2$. Since $\Sigma_2$ is the full shift over the finite alphabet $\{0,1\}$, every map from $\{0,\dots,n-1\}$ to $\{0,1\}$ occurs as an orbit block, and therefore \begin{align*} |\mathcal{L}_n(\Sigma_2)| = 2^n. \end{align*} We use the standard entropy formula for subshifts over a finite alphabet: if $X \subset A^{\mathbb{Z}}$ is a subshift over a finite alphabet $A$, then the topological entropy of the shift map on $X$ is \begin{align*} h_{\mathrm{top}}(\sigma|_X) = \lim_{n \to \infty} \frac{1}{n}\log |\mathcal{L}_n(X)|. \end{align*} The hypotheses of this formula hold because $\{0,1\}$ is finite and $\Sigma_2 \subset \{0,1\}^{\mathbb{Z}}$ is the full subshift. Hence \begin{align*} h_{\mathrm{top}}(\sigma) = \lim_{n \to \infty} \frac{1}{n}\log(2^n) = \log 2. \end{align*} [/step]

[step:Transfer the symbolic properties back to the horseshoe] The map $h: \Lambda \to \Sigma_2$ is a homeomorphism satisfying $h \circ F|_{\Lambda} = \sigma \circ h$. Therefore open sets, dense sets, and periodic orbits are carried bijectively between $F|_{\Lambda}$ and $\sigma$. It remains only to spell out the metric transfer for sensitivity. Let $d_{\Lambda}$ be the Euclidean metric restricted to $\Lambda$. Since $\Lambda$ is compact and $h: \Lambda \to \Sigma_2$ is continuous, $h$ is uniformly continuous. Hence there exists $\delta_{\Lambda} > 0$ such that, for all $x,y \in \Lambda$, \begin{align*} d_{\Lambda}(x,y) < \delta_{\Lambda} \implies d_{\Sigma}(h(x),h(y)) < 1/2. \end{align*} Equivalently, $d_{\Sigma}(h(x),h(y)) \ge 1/2$ implies $d_{\Lambda}(x,y) \ge \delta_{\Lambda}$. Given $x \in \Lambda$ and an open neighbourhood $W \subset \Lambda$ of $x$, the set $h(W)$ is an open neighbourhood of $h(x)$ in $\Sigma_2$. Sensitivity of $\sigma$ gives $b \in h(W)$ and $n \in \mathbb{N}$ such that \begin{align*} d_{\Sigma}(\sigma^n h(x),\sigma^n b) \ge 1/2. \end{align*} Set $y := h^{-1}(b) \in W$. Using $h \circ F|_{\Lambda} = \sigma \circ h$, we obtain \begin{align*} d_{\Sigma}(h(F^n(x)),h(F^n(y))) = d_{\Sigma}(\sigma^n h(x),\sigma^n b) \ge 1/2. \end{align*} Therefore $d_{\Lambda}(F^n(x),F^n(y)) \ge \delta_{\Lambda}$, so $F|_{\Lambda}$ has sensitive dependence on initial conditions. We use the standard conjugacy invariance theorem for topological entropy: if $X$ and $Y$ are compact topological spaces, $T: X \to X$ and $S: Y \to Y$ are continuous maps, and a homeomorphism $H: X \to Y$ satisfies $H \circ T = S \circ H$, then $h_{\mathrm{top}}(T)=h_{\mathrm{top}}(S)$. Its hypotheses hold here with $X=\Lambda$, $Y=\Sigma_2$, $T=F|_{\Lambda}$, $S=\sigma$, and $H=h$: compactness of $\Lambda$ is assumed, compactness of $\Sigma_2$ follows from compactness of the finite discrete alphabet $\{0,1\}$ and [Tychonoff's theorem](/theorems/953), $F|_{\Lambda}$ is continuous by hypothesis, $\sigma$ is continuous in the product topology, and $h$ is the assumed conjugating homeomorphism. Therefore \begin{align*} h_{\mathrm{top}}(F|_{\Lambda}) = h_{\mathrm{top}}(\sigma) = \log 2. \end{align*} The preceding steps show that $\sigma$ is topologically transitive, has dense periodic points, has sensitive dependence on initial conditions, and has entropy $\log 2$; hence $F|_{\Lambda}$ has the same four properties. [/step]

[step:Restrict the branch-crossing-only conclusion to the conjugate subsystem] Assume now that the trapping exclusion is not imposed, but that the branch-crossing construction still gives a nonempty compact invariant subsystem $\Lambda_0 \subset R$ and a homeomorphism \begin{align*} h_0: \Lambda_0 \to \Sigma_2 \end{align*} such that \begin{align*} h_0 \circ F|_{\Lambda_0} = \sigma \circ h_0. \end{align*} Repeating the conjugacy argument with $h_0$ in place of $h$ gives topological transitivity, dense periodic points, sensitive dependence, and \begin{align*} h_{\mathrm{top}}(F|_{\Lambda_0}) = \log 2 \end{align*} on the subsystem $\Lambda_0$. The entropy statement uses the same conjugacy invariance theorem, whose hypotheses hold because $\Lambda_0$ is compact by assumption, $F|_{\Lambda_0}$ is continuous on the invariant set under discussion, $\Sigma_2$ is compact, $\sigma$ is continuous, and $h_0$ is the assumed conjugating homeomorphism. Since the larger maximal invariant set in $R$ is not assumed to be conjugate to $\Sigma_2$, the argument gives no equality statement for all of its dynamics. It may contain additional invariant orbits or additional entropy beyond the horseshoe subsystem. This proves the asserted branch-crossing-only qualification and completes the proof. [/step]