Proof that equicontinous and suriective dynamical system is distal

Question

Let us have dynamical system $(X,f)$ (what means that $X$ is compact metric space with metric $d$ and $f: X \to X$ is continuous function). Moreover $f$ is surjective and equicontinous so when we take $x,y \in X$ we have $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall \ n \in \mathbb{N}: d(x,y) < \delta \Rightarrow d(f^n(x),f^n(y)) < \epsilon$. I want to show that such dynamical system is distal, so it does not contain proximal pair $\{x,y\}$ such that $x \ne y$ and $\lim \inf_{n \to \infty} d(f^n(x),f^n(y)) = 0$. My idea of proof is to take increasing sequence $n_k$ such that $\lim_{k \to \infty}d(f^{n_k}(x),f^{n_k}(y)) = 0$. This is of course equivalent to definition of proximal pair. Then using assumptions I want to show that $x = y$. Of course $\forall \epsilon > 0 \exists n_k \ d(f^{n_k}(x),f^{n_k}(y)) < \epsilon $. I have also suriectivity of $f$. I have problem though with formal proof. I wolud be grateful for any hint.

Answer 1

In addition to the above, suppose the family of functions $\{f^{-n}: n \in \mathbb{N}\}$ is equicontinuous on $X$.

Assume we have $x,y \in X$ such that $x\neq y$ and $\underset{n\to \infty}\liminf d(f^n(x),f^n(y)) = 0$. So there is a sequence of natural numbers $\{n_k\}$ such that $d\left(f^{n_k}(x),f^{n_k}(y)\right)<\delta$ for all $\delta>0$.

Let $x_k:=f^{n_k}(x)$ and $y_k:=f^{n_k}(y)$ $\, (k=1,2,\ldots)$.

Let $\delta>0$ be given. So we have $d\left(x_k,y_k\right)<\delta$ and $d\left(f^{-n_k}(x_k),f^{-n_k}(y_k)\right)=d(x,y)>0$.