[guided]The volume estimate says that every length-$n$ Bowen ball has measure comparable to $J_u^{-n}$. We now turn that geometric fact into an entropy computation. The bridge is simple: if sets of size about $J_u^{-n}$ cover a probability space of total measure $1$, then at least about $J_u^n$ of them are needed; conversely, a maximal separated family produces a cover, and disjoint smaller Bowen balls bound its size from above.
Fix $\varepsilon \in (0,\varepsilon_0]$. Let $F \subset \mathbb{T}^d$ be an $(n,\varepsilon)$-spanning set, meaning that
\begin{align*}
\mathbb{T}^d = \bigcup_{p \in F} B_n(p,\varepsilon).
\end{align*}
Since $\mu$ is a probability measure, $\mu(\mathbb{T}^d)=1$. Since measure is subadditive over countable unions and $F$ is finite for a minimal spanning set on the compact [metric space](/page/Metric%20Space) $\mathbb{T}^d$, we get
\begin{align*}
1=\mu(\mathbb{T}^d) \leq \sum_{p \in F}\mu(B_n(p,\varepsilon)).
\end{align*}
The upper Bowen-ball volume estimate gives $\mu(B_n(p,\varepsilon)) \leq c_2(\varepsilon)J_u^{-n}$ for every $p \in F$, with a constant independent of $p$ and $n$. Therefore
\begin{align*}
1 \leq |F| c_2(\varepsilon)J_u^{-n}.
\end{align*}
Solving this inequality for $|F|$ gives
\begin{align*}
|F| \geq c_2(\varepsilon)^{-1}J_u^n.
\end{align*}
Since $r_n(\varepsilon)$ is the smallest possible cardinality of such a spanning set, the same lower bound holds for $r_n(\varepsilon)$:
\begin{align*}
r_n(\varepsilon) \geq c_2(\varepsilon)^{-1}J_u^n.
\end{align*}
Taking logarithms, dividing by $n$, and letting $n \to \infty$, the fixed multiplicative constant disappears:
\begin{align*}
\liminf_{n\to\infty}\frac{1}{n}\log r_n(\varepsilon) \geq \log J_u.
\end{align*}
For the upper bound, choose $E_n \subset \mathbb{T}^d$ maximal with respect to the property that distinct points are separated in the Bowen metric at scale $\varepsilon/2$:
\begin{align*}
\rho_n(p,q)\geq \varepsilon/2 \quad \text{whenever } p,q \in E_n \text{ and } p \neq q.
\end{align*}
Maximality means that no point of $\mathbb{T}^d$ can be added while preserving this separation. Hence every point of $\mathbb{T}^d$ lies within $\rho_n$-distance less than $\varepsilon/2$ of some point of $E_n$, so $E_n$ is an $(n,\varepsilon/2)$-spanning set. In particular it is also an $(n,\varepsilon)$-spanning set, and therefore
\begin{align*}
r_n(\varepsilon) \leq |E_n|.
\end{align*}
Now distinct points of $E_n$ have $\rho_n$-distance at least $\varepsilon/2$. Therefore the smaller Bowen balls $B_n(p,\varepsilon/4)$ centered at points $p \in E_n$ are pairwise disjoint: if two such balls met, the triangle inequality for $\rho_n$ would put their centers within distance less than $\varepsilon/2$. Additivity of $\mu$ over pairwise disjoint measurable sets gives
\begin{align*}
1=\mu(\mathbb{T}^d) \geq \sum_{p \in E_n}\mu(B_n(p,\varepsilon/4)).
\end{align*}
The lower Bowen-ball volume estimate gives $\mu(B_n(p,\varepsilon/4)) \geq c_1(\varepsilon/4)J_u^{-n}$ for every $p \in E_n$. Hence
\begin{align*}
1 \geq |E_n| c_1(\varepsilon/4)J_u^{-n}.
\end{align*}
Solving for $|E_n|$ yields
\begin{align*}
|E_n| \leq c_1(\varepsilon/4)^{-1}J_u^n.
\end{align*}
Since $r_n(\varepsilon) \leq |E_n|$, we obtain
\begin{align*}
r_n(\varepsilon) \leq c_1(\varepsilon/4)^{-1}J_u^n.
\end{align*}
Taking logarithms, dividing by $n$, and letting $n \to \infty$ gives
\begin{align*}
\limsup_{n\to\infty}\frac{1}{n}\log r_n(\varepsilon) \leq \log J_u.
\end{align*}
The lower and upper estimates match, so
\begin{align*}
\lim_{n\to\infty}\frac{1}{n}\log r_n(\varepsilon)=\log J_u.
\end{align*}
Finally, the definition of topological entropy is obtained by taking the limit as the radius tends to $0$:
\begin{align*}
h_{\mathrm{top}}(f_A)=\lim_{\varepsilon \downarrow 0}\limsup_{n\to\infty}\frac{1}{n}\log r_n(\varepsilon)=\log J_u.
\end{align*}[/guided]