[proofplan] We compare the two definitions at finite time. A finite open cover with a Lebesgue number controls Bowen balls, so minimal spanning sets give subcovers of the iterated joins and hence $h_{\mathrm{cov}}(f)\le h_{\mathrm{top}}(f)$. Conversely, a finite cover by sets of diameter smaller than a Bowen scale separates the atoms of the iterated join, so separated sets inject into subcovers and hence $h_{\mathrm{top}}(f)\le h_{\mathrm{cov}}(f)$. Passing to logarithmic growth rates and then to the appropriate supremum or small-scale limit gives equality. [/proofplan]

[step:Fix the finite-time entropy notation] For each $n\in\mathbb N$, define the Bowen metric \begin{align*} d_n:X\times X\to [0,\infty),\qquad d_n(x,y)=\max_{0\le j\le n-1} d(f^j(x),f^j(y)). \end{align*} For $\varepsilon>0$, let $r_n(\varepsilon)$ denote the least cardinality of a set $E\subset X$ such that for every $x\in X$ there exists $e\in E$ with $d_n(x,e)<\varepsilon$. Such a set is called $(n,\varepsilon)$-spanning. Let $s_n(\varepsilon)$ denote the largest cardinality of a set $A\subset X$ such that for any distinct $x,y\in A$ one has $d_n(x,y)\ge \varepsilon$. Such a set is called $(n,\varepsilon)$-separated. Define the finite-scale spanning and separated growth rates by \begin{align*} R(\varepsilon)=\limsup_{n\to\infty}\frac{1}{n}\log r_n(\varepsilon) \end{align*} and \begin{align*} S(\varepsilon)=\limsup_{n\to\infty}\frac{1}{n}\log s_n(\varepsilon). \end{align*} The Bowen entropy is \begin{align*} h_{\mathrm{top}}(f)=\lim_{\varepsilon\downarrow 0} R(\varepsilon)=\lim_{\varepsilon\downarrow 0} S(\varepsilon), \end{align*} where the two limits agree by the standard spanning-separated comparison. For a finite open cover $\mathcal U$ of $X$, define its $n$-fold dynamical join by \begin{align*} \mathcal U_0^{n-1}=\bigvee_{j=0}^{n-1} f^{-j}\mathcal U. \end{align*} Thus $\mathcal U_0^{n-1}$ consists of all sets of the form \begin{align*} U_0\cap f^{-1}(U_1)\cap \cdots \cap f^{-(n-1)}(U_{n-1}), \end{align*} where $U_0,\dots,U_{n-1}\in\mathcal U$. Let $N(\mathcal V)$ denote the minimum cardinality of a subcover of a finite open cover $\mathcal V$ of $X$. The open-cover entropy of $\mathcal U$ is \begin{align*} h_{\mathrm{cov}}(f,\mathcal U)=\limsup_{n\to\infty}\frac{1}{n}\log N(\mathcal U_0^{n-1}), \end{align*} and the open-cover entropy of $f$ is \begin{align*} h_{\mathrm{cov}}(f)=\sup_{\mathcal U} h_{\mathrm{cov}}(f,\mathcal U), \end{align*} where the supremum is over all finite open covers $\mathcal U$ of $X$. [/step]

[step:Bound each cover entropy by a Bowen spanning entropy]Let $\mathcal U$ be a finite open cover of $X$. Since $X$ is compact and metric, the [Lebesgue number lemma](/theorems/952) applies to $\mathcal U$; more precisely, citing a result not yet in the wiki: Lebesgue Number Lemma. Choose a Lebesgue number $\lambda>0$ for $\mathcal U$ in the ball form: for each $z\in X$, there exists $U(z)\in\mathcal U$ such that \begin{align*} B_d(z,\lambda)\subset U(z). \end{align*} Fix $n\in\mathbb N$, and let $E\subset X$ be an $(n,\lambda)$-spanning set with $|E|=r_n(\lambda)$. For each $e\in E$ and each $j\in\{0,\dots,n-1\}$, choose $U_j(e)\in\mathcal U$ such that \begin{align*} B_d(f^j(e),\lambda)\subset U_j(e). \end{align*} Define the corresponding atom \begin{align*} A_e=U_0(e)\cap f^{-1}(U_1(e))\cap \cdots \cap f^{-(n-1)}(U_{n-1}(e)). \end{align*} Then $A_e\in\mathcal U_0^{n-1}$. We claim that the family $(A_e)_{e\in E}$ covers $X$. Let $x\in X$. Since $E$ is $(n,\lambda)$-spanning, there exists $e\in E$ such that $d_n(x,e)<\lambda$. Hence for every $j\in\{0,\dots,n-1\}$, \begin{align*} d(f^j(x),f^j(e))<\lambda. \end{align*} Therefore $f^j(x)\in B_d(f^j(e),\lambda)\subset U_j(e)$ for every $j$, and so $x\in A_e$. Thus \begin{align*} N(\mathcal U_0^{n-1})\le r_n(\lambda). \end{align*} Taking logarithms, dividing by $n$, and passing to $\limsup$ gives \begin{align*} h_{\mathrm{cov}}(f,\mathcal U)\le R(\lambda). \end{align*} Since $R(\lambda)\le \lim_{\varepsilon\downarrow 0}R(\varepsilon)=h_{\mathrm{top}}(f)$, we obtain \begin{align*} h_{\mathrm{cov}}(f,\mathcal U)\le h_{\mathrm{top}}(f). \end{align*}[/step]

[guided]The purpose of this step is to turn a Bowen spanning set into a subcover of the iterated cover. The link between the two notions is the Lebesgue number: it says that if an orbit point stays within $\lambda$ of a chosen centre at time $j$, then that orbit point lies in a single prescribed element of $\mathcal U$. Let $\mathcal U$ be a finite open cover of $X$. Because $X$ is compact and metric, we may apply the Lebesgue number lemma to $\mathcal U$; more precisely, citing a result not yet in the wiki: Lebesgue Number Lemma. Choose $\lambda>0$ such that for every $z\in X$ there is an element $U(z)\in\mathcal U$ satisfying \begin{align*} B_d(z,\lambda)\subset U(z). \end{align*} This is exactly the form needed for Bowen balls: the Bowen metric controls the ordinary metric after each iterate $f^j$. Fix $n\in\mathbb N$. Let $E\subset X$ be an $(n,\lambda)$-spanning set with minimal cardinality, so $|E|=r_n(\lambda)$. For each centre $e\in E$ and each time $j\in\{0,\dots,n-1\}$, the Lebesgue number property gives an [open set](/page/Open%20Set) $U_j(e)\in\mathcal U$ such that \begin{align*} B_d(f^j(e),\lambda)\subset U_j(e). \end{align*} Now define \begin{align*} A_e=U_0(e)\cap f^{-1}(U_1(e))\cap \cdots \cap f^{-(n-1)}(U_{n-1}(e)). \end{align*} This set belongs to $\mathcal U_0^{n-1}$ because it is one of the atoms of the join: membership in $A_e$ means that the point starts in $U_0(e)$, its first image lies in $U_1(e)$, and so on through time $n-1$. We verify that these atoms cover $X$. Take any $x\in X$. Since $E$ is $(n,\lambda)$-spanning, there exists $e\in E$ such that \begin{align*} d_n(x,e)<\lambda. \end{align*} By the definition of $d_n$, this means that for every $j\in\{0,\dots,n-1\}$, \begin{align*} d(f^j(x),f^j(e))<\lambda. \end{align*} Therefore $f^j(x)\in B_d(f^j(e),\lambda)$, and the chosen containment of the ball gives $f^j(x)\in U_j(e)$. This holds for every $j$, so $x\in A_e$. Thus $(A_e)_{e\in E}$ is a subcover of $\mathcal U_0^{n-1}$ with at most $|E|=r_n(\lambda)$ elements. Hence \begin{align*} N(\mathcal U_0^{n-1})\le r_n(\lambda). \end{align*} Taking logarithms and dividing by $n$ preserves the inequality because logarithm is increasing. Passing to $\limsup_{n\to\infty}$ yields \begin{align*} h_{\mathrm{cov}}(f,\mathcal U)\le R(\lambda). \end{align*} Finally, the Bowen entropy is the small-scale limit of $R(\varepsilon)$ as $\varepsilon\downarrow 0$, so the fixed-scale quantity $R(\lambda)$ is bounded above by $h_{\mathrm{top}}(f)$. Therefore \begin{align*} h_{\mathrm{cov}}(f,\mathcal U)\le h_{\mathrm{top}}(f). \end{align*}[/guided]

[step:Take the supremum over finite covers to obtain the first inequality] The preceding step holds for every finite open cover $\mathcal U$ of $X$. Taking the supremum over all such $\mathcal U$ gives \begin{align*} h_{\mathrm{cov}}(f)=\sup_{\mathcal U}h_{\mathrm{cov}}(f,\mathcal U)\le h_{\mathrm{top}}(f). \end{align*} [/step]

[step:Bound separated sets by subcovers of a fine open cover]Fix $\varepsilon>0$. Since $X$ is compact and metric, choose a finite open cover $\mathcal U_\varepsilon$ of $X$ such that every $U\in\mathcal U_\varepsilon$ has diameter strictly smaller than $\varepsilon$. For example, finitely many open balls of radius $\varepsilon/2$ suffice. Fix $n\in\mathbb N$, and let $A\subset X$ be an $(n,\varepsilon)$-separated set with $|A|=s_n(\varepsilon)$. We claim that every atom of $\mathcal U_{\varepsilon,0}^{n-1}$ contains at most one point of $A$. Indeed, suppose $x,y\in A$ lie in the same atom \begin{align*} C=U_0\cap f^{-1}(U_1)\cap \cdots \cap f^{-(n-1)}(U_{n-1}), \end{align*} where $U_0,\dots,U_{n-1}\in\mathcal U_\varepsilon$. Then for every $j\in\{0,\dots,n-1\}$, both $f^j(x)$ and $f^j(y)$ belong to $U_j$. Since $\operatorname{diam}(U_j)<\varepsilon$, \begin{align*} d(f^j(x),f^j(y))<\varepsilon. \end{align*} Taking the maximum over $j$ gives \begin{align*} d_n(x,y)<\varepsilon. \end{align*} This contradicts $(n,\varepsilon)$-separation unless $x=y$. Let $\mathcal W_n\subset\mathcal U_{\varepsilon,0}^{n-1}$ be a subcover with $|\mathcal W_n|=N(\mathcal U_{\varepsilon,0}^{n-1})$. Since each member of $\mathcal W_n$ contains at most one point of $A$ and $\mathcal W_n$ covers $A$, we have \begin{align*} s_n(\varepsilon)=|A|\le N(\mathcal U_{\varepsilon,0}^{n-1}). \end{align*} Taking logarithms, dividing by $n$, and passing to $\limsup$ gives \begin{align*} S(\varepsilon)\le h_{\mathrm{cov}}(f,\mathcal U_\varepsilon)\le h_{\mathrm{cov}}(f). \end{align*}[/step]

[guided]Now we prove the reverse comparison. This time the cover must be chosen from the metric scale. Fix $\varepsilon>0$. Since $X$ is compact, the open balls $B_d(x,\varepsilon/2)$ with $x\in X$ admit a finite subcover. Let $\mathcal U_\varepsilon$ be such a finite subcover. Every member of $\mathcal U_\varepsilon$ has diameter strictly smaller than $\varepsilon$, because if $a,b\in B_d(x,\varepsilon/2)$, then the triangle inequality gives \begin{align*} d(a,b)\le d(a,x)+d(x,b)<\varepsilon. \end{align*} Fix $n\in\mathbb N$, and let $A\subset X$ be an $(n,\varepsilon)$-separated set of maximal cardinality, so $|A|=s_n(\varepsilon)$. The key point is that a single atom of the join cannot contain two distinct points of $A$. To see this, take an atom \begin{align*} C=U_0\cap f^{-1}(U_1)\cap \cdots \cap f^{-(n-1)}(U_{n-1}), \end{align*} with $U_0,\dots,U_{n-1}\in\mathcal U_\varepsilon$. Suppose $x,y\in A\cap C$. For each $j\in\{0,\dots,n-1\}$, membership in $C$ gives $f^j(x)\in U_j$ and $f^j(y)\in U_j$. Because $\operatorname{diam}(U_j)<\varepsilon$, we get \begin{align*} d(f^j(x),f^j(y))<\varepsilon. \end{align*} Since this holds for every $j$, the definition of the Bowen metric gives \begin{align*} d_n(x,y)=\max_{0\le j\le n-1}d(f^j(x),f^j(y))<\varepsilon. \end{align*} But $A$ is $(n,\varepsilon)$-separated, so distinct points of $A$ must have $d_n$-distance at least $\varepsilon$. Therefore the only possible conclusion is $x=y$. Hence each atom contains at most one point of $A$. Let $\mathcal W_n$ be a subcover of $\mathcal U_{\varepsilon,0}^{n-1}$ with minimal cardinality: \begin{align*} |\mathcal W_n|=N(\mathcal U_{\varepsilon,0}^{n-1}). \end{align*} The family $\mathcal W_n$ covers all of $X$, hence it covers $A$. Since no member of $\mathcal W_n$ can contain more than one point of $A$, the number of points in $A$ is at most the number of sets in $\mathcal W_n$. Thus \begin{align*} s_n(\varepsilon)=|A|\le N(\mathcal U_{\varepsilon,0}^{n-1}). \end{align*} Taking logarithms, dividing by $n$, and then passing to $\limsup_{n\to\infty}$ gives \begin{align*} S(\varepsilon)\le h_{\mathrm{cov}}(f,\mathcal U_\varepsilon). \end{align*} Since $h_{\mathrm{cov}}(f)$ is the supremum of $h_{\mathrm{cov}}(f,\mathcal U)$ over all finite open covers $\mathcal U$, we also have \begin{align*} h_{\mathrm{cov}}(f,\mathcal U_\varepsilon)\le h_{\mathrm{cov}}(f). \end{align*} Combining the two inequalities yields \begin{align*} S(\varepsilon)\le h_{\mathrm{cov}}(f). \end{align*}[/guided]

[step:Let the Bowen scale shrink to obtain the second inequality] The estimate from the preceding step holds for every $\varepsilon>0$. Therefore \begin{align*} h_{\mathrm{top}}(f)=\lim_{\varepsilon\downarrow 0}S(\varepsilon)\le h_{\mathrm{cov}}(f). \end{align*} Together with the inequality $h_{\mathrm{cov}}(f)\le h_{\mathrm{top}}(f)$ already proved, this gives \begin{align*} h_{\mathrm{top}}(f)=h_{\mathrm{cov}}(f). \end{align*} [/step]