 Solution

Use the separated-set definition of topological entropy. For a compatible metric $d$ and $n\ge 1$, define the Bowen metric and separated-set count byFor fixed $n$, the metric $d_n$ generates the original topology: it includes the $k=0$ term and all finitely many iterations of a map are continuous. Thus its space is compact and $s_n(d,ϵ)$ is finite. With natural logarithms, setThe last limit exists, possibly with value infinity, because decreasing $ϵ$ can only increase separated-set counts.

Let $d^{\prime }$ be another compatible metric. The identity $(X,d)\to (X,d^{\prime })$ is a continuous function on a compact domain, hence uniformly continuous. For every $ϵ>0$, choose $\delta >0$ such that $d(x,y)<\delta$ implies $d^{\prime }(x,y)<ϵ$. The same implication at every iterate givesBy its contrapositive, every $ϵ$-separated set for $d_n^{\prime }$ is $\delta$-separated for $d_n$. Consequently $s_n(d^{\prime },ϵ)\le s_n(d,\delta )$ for every $n$, andTaking $ϵ\downarrow 0$ gives $h_{d^{\prime }}(f)\le h_d(f)$. Interchanging the two compatible metrics gives the reverse inequality. ThereforeCompactness is used for uniform continuity, not merely for pointwise equivalence of the two metrics.