[proofplan] We prove the entropy lower bound directly from separated sets. First we recall the separated-set definition of topological entropy and show that the full two-shift has entropy at least $\log 2$ by constructing $2^m$ orbit segments separated for $m$ iterates. Then we pull separated sets back through the factor map $\pi$, use the inclusion $\Lambda \subset X$ to compare the restricted system with the ambient system, and finally compare $h_{\mathrm{top}}(f^N)$ with $h_{\mathrm{top}}(f)$ through the Bowen metrics for iterates. [/proofplan]

[step:Set up separated-set notation for entropy] Let $(Y,\rho)$ be a compact [metric space](/page/Metric%20Space), and let $S: Y \to Y$ be continuous. For each $m \in \mathbb{N}$, define the Bowen metric $\rho_m^S: Y \times Y \to [0,\infty)$ by \begin{align*} \rho_m^S(y,z) := \max_{0 \leq j < m} \rho(S^j(y),S^j(z)). \end{align*} For $\varepsilon > 0$, a finite set $E \subset Y$ is called $(m,\varepsilon)$-separated for $S$ if $\rho_m^S(y,z) > \varepsilon$ whenever $y,z \in E$ and $y \neq z$. Let $s_m(S,\varepsilon)$ denote the maximal cardinality of an $(m,\varepsilon)$-separated subset of $Y$. The topological entropy of $S$ is \begin{align*} h_{\mathrm{top}}(S) := \sup_{\varepsilon > 0} \limsup_{m \to \infty} \frac{1}{m}\log s_m(S,\varepsilon). \end{align*} [/step]

[step:Construct exponentially many separated orbits in the full two-shift] Equip $\Sigma_2$ with the compatible metric $\rho_\Sigma: \Sigma_2 \times \Sigma_2 \to [0,\infty)$ defined by \begin{align*} \rho_\Sigma(a,b) := \sum_{k \in \mathbb{Z}} 2^{-|k|-2}|a_k-b_k|, \end{align*} where $a=(a_k)_{k \in \mathbb{Z}}$ and $b=(b_k)_{k \in \mathbb{Z}}$. Set $\varepsilon_0 := 1/8$. Fix $m \in \mathbb{N}$. For each word $w=(w_0,\dots,w_{m-1}) \in \{0,1\}^m$, define a sequence $a^w \in \Sigma_2$ by setting $a^w_k=w_k$ for $0\leq k<m$ and setting $a^w_k=0$ for $k<0$ or $k\geq m$. Let $E_m := \{a^w : w \in \{0,1\}^m\}$. Then $|E_m|=2^m$. If $w,v \in \{0,1\}^m$ are distinct, choose an index $j \in \{0,\dots,m-1\}$ such that $w_j \neq v_j$. Since $(\sigma^j a^w)_0 = w_j$ and $(\sigma^j a^v)_0 = v_j$, the zeroth-coordinate term in the metric gives \begin{align*} \rho_\Sigma(\sigma^j a^w,\sigma^j a^v) \geq 2^{-2}|w_j-v_j| = \frac{1}{4}. \end{align*} Therefore $\rho_m^\sigma(a^w,a^v) > \varepsilon_0$. Hence $E_m$ is $(m,\varepsilon_0)$-separated for $\sigma$, and so \begin{align*} s_m(\sigma,\varepsilon_0) \geq 2^m. \end{align*} It follows that \begin{align*} h_{\mathrm{top}}(\sigma) \geq \limsup_{m \to \infty}\frac{1}{m}\log(2^m) = \log 2. \end{align*} [/step]

[step:Pull separated sets back through the factor map]Define $T: \Lambda \to \Lambda$ by $T := f^N|_\Lambda$. The hypothesis $f^N(\Lambda) \subset \Lambda$ makes $T$ a continuous self-map of the compact metric space $\Lambda$, equipped with the restricted metric $d|_{\Lambda \times \Lambda}$. The factor relation is \begin{align*} \pi \circ T = \sigma \circ \pi. \end{align*} Fix $\eta > 0$. Since $\Lambda$ is compact and $\pi: \Lambda \to \Sigma_2$ is continuous, $\pi$ is uniformly continuous. Hence there exists $\delta > 0$ such that for all $x,y \in \Lambda$, \begin{align*} d(x,y) < \delta \implies \rho_\Sigma(\pi(x),\pi(y)) < \eta. \end{align*} We claim that $s_m(T,\delta) \geq s_m(\sigma,\eta)$ for every $m \in \mathbb{N}$. Let $A \subset \Sigma_2$ be an $(m,\eta)$-separated set for $\sigma$. Since $\pi$ is surjective, for each $a \in A$ choose $x_a \in \Lambda$ with $\pi(x_a)=a$, and define $B := \{x_a : a \in A\}$. If $a,b \in A$ are distinct, there exists $j \in \{0,\dots,m-1\}$ such that \begin{align*} \rho_\Sigma(\sigma^j a,\sigma^j b) > \eta. \end{align*} Using the factor relation, this becomes \begin{align*} \rho_\Sigma(\pi(T^j x_a),\pi(T^j x_b)) > \eta. \end{align*} By the contrapositive of [uniform continuity](/page/Uniform%20Continuity), $d(T^j x_a,T^j x_b) \geq \delta$, and after replacing $\delta$ by any smaller positive number if necessary, the set $B$ is $(m,\delta')$-separated for every $0<\delta'<\delta$. Thus \begin{align*} h_{\mathrm{top}}(T) \geq h_{\mathrm{top}}(\sigma) \geq \log 2. \end{align*}[/step]

[guided]The factor map goes from the subsystem $(\Lambda,T)$ onto the symbolic system $(\Sigma_2,\sigma)$. Entropy cannot increase when passing to a factor, but here we verify exactly what that means using separated sets. Let $\eta > 0$ be a separation scale in the shift space $\Sigma_2$. Since $\Lambda$ is compact and $\pi: \Lambda \to \Sigma_2$ is continuous, $\pi$ is uniformly continuous. Therefore there exists $\delta > 0$ such that whenever $x,y \in \Lambda$ satisfy $d(x,y)<\delta$, their images satisfy \begin{align*} \rho_\Sigma(\pi(x),\pi(y)) < \eta. \end{align*} Equivalently, if $\rho_\Sigma(\pi(x),\pi(y)) \geq \eta$, then $d(x,y) \geq \delta$. Now fix $m \in \mathbb{N}$ and let $A \subset \Sigma_2$ be an $(m,\eta)$-separated set for the shift $\sigma$. Because $\pi$ is surjective, every point $a \in A$ has at least one preimage in $\Lambda$. Choose one such point and call it $x_a \in \Lambda$, so $\pi(x_a)=a$. Define \begin{align*} B := \{x_a : a \in A\}. \end{align*} The set $B$ has the same cardinality as $A$. We must show that the points in $B$ are separated under the dynamics of $T$. Take distinct $a,b \in A$. Since $A$ is $(m,\eta)$-separated, there exists an index $j \in \{0,\dots,m-1\}$ such that \begin{align*} \rho_\Sigma(\sigma^j a,\sigma^j b) > \eta. \end{align*} The intertwining relation $\pi \circ T = \sigma \circ \pi$ implies, by induction on $j$, that $\pi \circ T^j = \sigma^j \circ \pi$. Therefore \begin{align*} \sigma^j a = \sigma^j \pi(x_a) = \pi(T^j x_a) \end{align*} and \begin{align*} \sigma^j b = \sigma^j \pi(x_b) = \pi(T^j x_b). \end{align*} Substituting these identities gives \begin{align*} \rho_\Sigma(\pi(T^j x_a),\pi(T^j x_b)) > \eta. \end{align*} By the contrapositive of uniform continuity, $d(T^j x_a,T^j x_b) \geq \delta$. Hence the Bowen distance between $x_a$ and $x_b$ for the map $T$ is at least $\delta$. If one uses the strict $m$-step Bowen separated convention, this proves that $B$ is $(m,\delta')$-separated for every $0<\delta'<\delta$. Thus every separated set in the shift has a separated lift in $\Lambda$ with the same cardinality. Passing to maximal cardinalities, then to exponential growth rates, gives \begin{align*} h_{\mathrm{top}}(T) \geq h_{\mathrm{top}}(\sigma). \end{align*} From the previous step, $h_{\mathrm{top}}(\sigma) \geq \log 2$, so \begin{align*} h_{\mathrm{top}}(T) \geq \log 2. \end{align*}[/guided]

[step:Compare the restricted invariant system with the ambient system] For each $m \in \mathbb{N}$ and $\varepsilon > 0$, every $(m,\varepsilon)$-separated subset of $\Lambda$ for $T=f^N|_\Lambda$ is also an $(m,\varepsilon)$-separated subset of $X$ for $f^N$, because the metric on $\Lambda$ is the restriction of $d$ and $T^j(x)=f^{Nj}(x)$ for every $x \in \Lambda$ and $0 \leq j < m$. Therefore \begin{align*} s_m(T,\varepsilon) \leq s_m(f^N,\varepsilon). \end{align*} Taking exponential growth rates and then the supremum over $\varepsilon>0$ yields \begin{align*} h_{\mathrm{top}}(f^N) \geq h_{\mathrm{top}}(T). \end{align*} Consequently, \begin{align*} h_{\mathrm{top}}(f^N) \geq \log 2. \end{align*} [/step]

[step:Compare the entropy of an iterate with the entropy of the original map] Fix $\varepsilon > 0$ and $m \in \mathbb{N}$. If $E \subset X$ is $(m,\varepsilon)$-separated for $f^N$, then for every distinct $x,y \in E$ there is an index $j \in \{0,\dots,m-1\}$ such that \begin{align*} d(f^{Nj}(x),f^{Nj}(y)) > \varepsilon. \end{align*} Since $Nj \in \{0,\dots,Nm-1\}$, the same pair is separated within the first $Nm$ iterates of $f$. Hence $E$ is $(Nm,\varepsilon)$-separated for $f$, and therefore \begin{align*} s_m(f^N,\varepsilon) \leq s_{Nm}(f,\varepsilon). \end{align*} It follows that \begin{align*} \limsup_{m \to \infty}\frac{1}{m}\log s_m(f^N,\varepsilon) \leq N \limsup_{k \to \infty}\frac{1}{k}\log s_k(f,\varepsilon). \end{align*} Taking the supremum over $\varepsilon>0$ gives \begin{align*} h_{\mathrm{top}}(f^N) \leq N h_{\mathrm{top}}(f). \end{align*} [/step]

[step:Combine the inequalities and divide by the positive iterate] From the previous two steps, \begin{align*} \log 2 \leq h_{\mathrm{top}}(f^N) \leq N h_{\mathrm{top}}(f). \end{align*} Since $N \in \mathbb{N}$, we have $N>0$, so division by $N$ gives \begin{align*} h_{\mathrm{top}}(f) \geq \frac{\log 2}{N}. \end{align*} This is the desired lower bound. [/step]