[proofplan] We prove the two entropy inequalities directly using separated orbit segments. First, separated sets for the subsystem $(\Lambda,f|_\Lambda)$ are also separated sets for $(X,f)$, giving $h_{\mathrm{top}}(f)\ge h_{\mathrm{top}}(f|_\Lambda)$. Second, the factor map $\pi:\Lambda\to\Sigma_k$ lets us lift separated orbit segments from the full shift to $\Lambda$ by [uniform continuity](/page/Uniform%20Continuity). Finally, the full $k$-shift has at least $k^n$ orbit segments separated for $n$ iterates, so its entropy is at least $\log k$. [/proofplan]

[step:Compare the entropy of $f$ with the entropy of its invariant restriction]Let $d_\Lambda:\Lambda\times\Lambda\to[0,\infty)$ denote the restricted metric, defined by $d_\Lambda(x,y)=d(x,y)$. For $n\in\mathbb{N}$ and $\varepsilon>0$, define $s_X(n,\varepsilon)$ to be the maximal cardinality of an $(n,\varepsilon)$-separated subset of $X$ for $f$, and define $s_\Lambda(n,\varepsilon)$ to be the maximal cardinality of an $(n,\varepsilon)$-separated subset of $\Lambda$ for $g=f|_\Lambda$. We use the separated-set definition \begin{align*} h_{\mathrm{top}}(f)=\sup_{\varepsilon>0}\limsup_{n\to\infty}\frac{1}{n}\log s_X(n,\varepsilon) \end{align*} and the same definition for $h_{\mathrm{top}}(g)$ with $s_\Lambda(n,\varepsilon)$ in place of $s_X(n,\varepsilon)$. If $E\subset\Lambda$ is $(n,\varepsilon)$-separated for $g$, then for any distinct $x,y\in E$ there exists an index $j\in\{0,\dots,n-1\}$ such that \begin{align*} d_\Lambda(g^j(x),g^j(y))>\varepsilon. \end{align*} Since $g^j(x)=f^j(x)$ and $d_\Lambda$ is the restriction of $d$, this is exactly \begin{align*} d(f^j(x),f^j(y))>\varepsilon. \end{align*} Thus $E$ is also $(n,\varepsilon)$-separated for $f$ in $X$. Hence \begin{align*} s_X(n,\varepsilon)\ge s_\Lambda(n,\varepsilon) \end{align*} for every $n\in\mathbb{N}$ and every $\varepsilon>0$. Taking exponential growth rates and then the supremum over $\varepsilon>0$ gives \begin{align*} h_{\mathrm{top}}(f)\ge h_{\mathrm{top}}(g)=h_{\mathrm{top}}(f|_\Lambda). \end{align*}[/step]

[guided]The restriction $g=f|_\Lambda$ is a well-defined continuous map $\Lambda\to\Lambda$ because $f(\Lambda)\subseteq\Lambda$. We compare entropy using the separated-set definition. For each $n\in\mathbb{N}$ and $\varepsilon>0$, let $s_X(n,\varepsilon)$ denote the largest size of a subset $E\subset X$ such that any two distinct points $x,y\in E$ become $\varepsilon$-apart during the first $n$ iterates of $f$. Similarly, let $s_\Lambda(n,\varepsilon)$ denote the corresponding quantity inside $\Lambda$ for the restricted map $g$. Now suppose $E\subset\Lambda$ is $(n,\varepsilon)$-separated for $g$. This means that for any distinct $x,y\in E$, there is some $j\in\{0,\dots,n-1\}$ with \begin{align*} d_\Lambda(g^j(x),g^j(y))>\varepsilon. \end{align*} Because $g=f|_\Lambda$, we have $g^j(x)=f^j(x)$ for every $x\in\Lambda$ and every $j\ge0$. Also, $d_\Lambda$ is just the metric $d$ restricted to $\Lambda\times\Lambda$. Therefore the same inequality becomes \begin{align*} d(f^j(x),f^j(y))>\varepsilon. \end{align*} So every separated orbit family inside $\Lambda$ is automatically a separated orbit family in the larger space $X$. Therefore \begin{align*} s_X(n,\varepsilon)\ge s_\Lambda(n,\varepsilon) \end{align*} for all $n$ and $\varepsilon$. Passing to the limiting exponential growth rate in $n$, and then taking the supremum over separation scales $\varepsilon>0$, yields \begin{align*} h_{\mathrm{top}}(f)\ge h_{\mathrm{top}}(f|_\Lambda). \end{align*}[/guided]

[step:Lift separated orbit segments from the full shift through the factor map]Let $\rho:\Sigma_k\times\Sigma_k\to[0,\infty)$ be any metric compatible with the [product topology](/page/Product%20Topology) on $\Sigma_k$. Because the one-symbol cylinders \begin{align*} C_a:=\{\omega\in\Sigma_k:\omega_0=a\},\qquad a\in\{1,\dots,k\}, \end{align*} are pairwise disjoint compact sets, define \begin{align*} \varepsilon_0:=\frac{1}{2}\min_{a\ne b}\operatorname{dist}_\rho(C_a,C_b)>0. \end{align*} Since $\pi:\Lambda\to\Sigma_k$ is continuous and $\Lambda$ is compact, $\pi$ is uniformly continuous. Hence there exists $\delta>0$ such that for all $x,y\in\Lambda$, \begin{align*} d_\Lambda(x,y)<\delta \implies \rho(\pi(x),\pi(y))<\varepsilon_0. \end{align*} For each word $w=(w_0,\dots,w_{n-1})\in\{1,\dots,k\}^n$, choose a sequence $\omega_w\in\Sigma_k$ whose coordinates $0,\dots,n-1$ are $w_0,\dots,w_{n-1}$. Since $\pi$ is surjective, choose $x_w\in\Lambda$ with $\pi(x_w)=\omega_w$. If $w\ne v$, choose $j\in\{0,\dots,n-1\}$ with $w_j\ne v_j$. Then $\sigma^j(\omega_w)$ and $\sigma^j(\omega_v)$ lie in different one-symbol cylinders, so \begin{align*} \rho(\sigma^j(\omega_w),\sigma^j(\omega_v))\ge 2\varepsilon_0>\varepsilon_0. \end{align*} Using the semiconjugacy relation $\pi\circ g=\sigma\circ\pi$, we get \begin{align*} \rho(\pi(g^j(x_w)),\pi(g^j(x_v)))=\rho(\sigma^j(\omega_w),\sigma^j(\omega_v))>\varepsilon_0. \end{align*} By the contrapositive of the uniform-continuity implication, this gives \begin{align*} d_\Lambda(g^j(x_w),g^j(x_v))\ge\delta. \end{align*} Define \begin{align*} \delta_1:=\frac{\delta}{2}. \end{align*} Then $d_\Lambda(g^j(x_w),g^j(x_v))\ge\delta>\delta_1$. Thus the set \begin{align*} E_n:=\{x_w:w\in\{1,\dots,k\}^n\} \end{align*} is $(n,\delta_1)$-separated for $g$. The points in $E_n$ are distinct, because $x_w=x_v$ would imply $\omega_w=\pi(x_w)=\pi(x_v)=\omega_v$, contradicting the choice of words when $w\ne v$. Hence $E_n$ has cardinality $k^n$, and therefore \begin{align*} s_\Lambda(n,\delta_1)\ge k^n \end{align*} for every $n\in\mathbb{N}$.[/step]

[guided]The purpose of this step is to transfer symbolic separation from the full shift back to $\Lambda$. Let $\rho:\Sigma_k\times\Sigma_k\to[0,\infty)$ be a metric compatible with the product topology. For each symbol $a\in\{1,\dots,k\}$, define the one-symbol cylinder \begin{align*} C_a:=\{\omega\in\Sigma_k:\omega_0=a\}. \end{align*} Because $\Sigma_k$ is compact and each $C_a$ is closed, each $C_a$ is compact. The cylinders $C_a$ and $C_b$ are disjoint whenever $a\ne b$, so compactness in the [metric space](/page/Metric%20Space) $(\Sigma_k,\rho)$ gives $\operatorname{dist}_\rho(C_a,C_b)>0$. Since there are only finitely many pairs, the number \begin{align*} \varepsilon_0:=\frac{1}{2}\min_{a\ne b}\operatorname{dist}_\rho(C_a,C_b) \end{align*} is positive. This means that two sequences whose zeroth symbols differ are at least $2\varepsilon_0$ apart in the metric $\rho$. The map $\pi:\Lambda\to\Sigma_k$ is continuous and $\Lambda$ is compact, so $\pi$ is uniformly continuous. Therefore there exists $\delta>0$ such that for all $x,y\in\Lambda$, \begin{align*} d_\Lambda(x,y)<\delta \implies \rho(\pi(x),\pi(y))<\varepsilon_0. \end{align*} We will use the contrapositive of this implication: if $\rho(\pi(x),\pi(y))\ge\varepsilon_0$, then $d_\Lambda(x,y)\ge\delta$. The strict separated-set definition requires a strict lower bound, so we set \begin{align*} \delta_1:=\frac{\delta}{2}. \end{align*} Then any pair with $d_\Lambda(x,y)\ge\delta$ is certainly more than $\delta_1$ apart. Fix $n\in\mathbb{N}$. For each word $w=(w_0,\dots,w_{n-1})\in\{1,\dots,k\}^n$, choose a sequence $\omega_w\in\Sigma_k$ whose coordinates $0,\dots,n-1$ are $w_0,\dots,w_{n-1}$. This choice is possible both for the one-sided shift indexed by $\mathbb{N}_0$ and for the two-sided shift indexed by $\mathbb{Z}$. Since $\pi$ is surjective, choose $x_w\in\Lambda$ such that $\pi(x_w)=\omega_w$. Now take two distinct words $w\ne v$. Choose $j\in\{0,\dots,n-1\}$ such that $w_j\ne v_j$. After applying $\sigma^j$, the sequences $\sigma^j(\omega_w)$ and $\sigma^j(\omega_v)$ have different zeroth symbols. Hence they lie in two different one-symbol cylinders, so \begin{align*} \rho(\sigma^j(\omega_w),\sigma^j(\omega_v))\ge 2\varepsilon_0>\varepsilon_0. \end{align*} The semiconjugacy identity $\pi\circ g=\sigma\circ\pi$ gives, by induction on $j$, \begin{align*} \pi(g^j(x_w))=\sigma^j(\pi(x_w))=\sigma^j(\omega_w) \end{align*} and likewise $\pi(g^j(x_v))=\sigma^j(\omega_v)$. Therefore \begin{align*} \rho(\pi(g^j(x_w)),\pi(g^j(x_v)))=\rho(\sigma^j(\omega_w),\sigma^j(\omega_v))>\varepsilon_0. \end{align*} By the contrapositive of uniform continuity, this implies \begin{align*} d_\Lambda(g^j(x_w),g^j(x_v))\ge\delta>\delta_1. \end{align*} Thus the set \begin{align*} E_n:=\{x_w:w\in\{1,\dots,k\}^n\} \end{align*} is $(n,\delta_1)$-separated for $g$. Also, $x_w=x_v$ would force $\omega_w=\pi(x_w)=\pi(x_v)=\omega_v$, which is impossible for distinct words because the chosen sequences differ in at least one of the first $n$ coordinates. Hence $|E_n|=k^n$, and so \begin{align*} s_\Lambda(n,\delta_1)\ge k^n \end{align*} for every $n\in\mathbb{N}$.[/guided]

[step:Convert the lifted separated sets into the entropy lower bound] From the previous step, for the fixed number $\delta_1>0$, \begin{align*} s_\Lambda(n,\delta_1)\ge k^n \end{align*} for every $n\in\mathbb{N}$. Hence \begin{align*} \limsup_{n\to\infty}\frac{1}{n}\log s_\Lambda(n,\delta_1)\ge \lim_{n\to\infty}\frac{1}{n}\log(k^n)=\log k. \end{align*} Taking the supremum over all positive separation scales in the definition of topological entropy gives \begin{align*} h_{\mathrm{top}}(f|_\Lambda)=h_{\mathrm{top}}(g)\ge \log k. \end{align*} Combining this with the first step yields \begin{align*} h_{\mathrm{top}}(f)\ge h_{\mathrm{top}}(f|_\Lambda)\ge \log k. \end{align*} Since $k\ge2$, we have $\log k>0$, and therefore $h_{\mathrm{top}}(f)>0$. [/step]