I'm dealing with the question proving the Bowen metrics induce the same topology. More specifically, given $(X,T)$ be a topological dynamical systems equipped with a metric $d$ ($X$ is compact, $T$ is continuous). For each $n \ge 1$, define $$d_n(x,y) = \max \{d(T^k(x),T^k(y)) : k = 0,1,\ldots,n-1 \}.$$ We call $d_n$ the Bowen metrics on $X$. Also, let $B_{n,\epsilon}(x)$ denote the open ball about $x$ of radius $\epsilon$ in the metric topology induced by $d_n$.
Now, our aim is to show that $d_n$'s induced the same topology on $X$.
My attempt: Let $\mathcal{T}$ be the topology induced by $d$, and for $n \ge 2$, let $\mathcal{T}_n$ be the topology induced by $d_n$. First, we can show that $\mathcal{T} \subset \mathcal{T}_n$. Indeed, let $A \in \mathcal{T}$. Then, there is some $\epsilon > 0$ such that $B_{1,\epsilon}(x) \subset A$. Let $y \in B_{n,\epsilon}(x)$, so $d_n(x,y) < \epsilon$. Since $$d(z,w) \le d_n(z,w), \forall z,w \in X,$$ we have $d(x,y) < \epsilon$. Hence, $y \in B_{1,\epsilon}(x)$. This implies $$B_{n,\epsilon}(x) \subset B_{1,\epsilon}(x) \subset A.$$ Therefore, $A \in \mathcal{T}_n$, which leads to $\mathcal{T} \subset \mathcal{T}_n$.
For the converse, I need to show $\mathcal{T}_n \subset \mathcal{T}$. Let $A \in \mathcal{T}_n$. Then, there is some $\epsilon > 0$ such that $B_{n,\epsilon}(x) \subset A$. I need to choose $\epsilon' > 0$ such that $$B_{1,\epsilon'}(x) \subset B_{n,\epsilon}(x) \subset A.$$ However, I am stuck at that point. Are there any suggestions?
Any help is highly appreciated. Thank you!