Equivalence of the definitions of exactness and mixing

Question

Let $f:X \to X$ be a continuous map, where $X$ is a compact metric space. We say that $f$ is expanding if there are constants $\lambda >1$ and $\delta_0 > 0$ such that, for all $x, y\in X$, $d(f(x), f(y)) \ge \lambda d(x, y)$ whenever $d(x, y) \le \delta_0$.

Reading this article and searching the internet, it seems that for an expanding map the definitions of exactness and mixing are equivalent.

1. (Mixing) if for every pair of open and not-empty sets $U, V \subset X$, there exists an $n_0 \in \mathbb{N}$ such that $f^n(U) \cap V \neq \emptyset$ for all $n \ge n_0$.
2. (Exact or Locally Eventually Onto) if for every open and not-empty $U \subset X$, there exists $n_0\in\mathbb{N}$ such that $f^{n_0}(U) = X$.

I found an exercise that asks to prove that for an expanding map one has $1\Rightarrow 2$, but I couldn't complete the exercise. The case $2\Rightarrow 1$ seems trivial. Because if exists $n_0$, $f^{n_0}(U)=X$, for any $V\subset X$ the intersection will not be empty. In this case it was not necessary for the map to be expanding.

So the problem for me is at $1\Rightarrow 2$. Any reference for the proof? Or a help to conclude...

For clarification, $f^n(x)=f\circ f \circ \cdots \circ f(x)$ an iterative map. Here we are using the topological definition of mixing and exactness.

The exercise is from the book "Foundations of Ergodic Theory" by Viana and Oliveira, where an extra hypothesis is needed in the definition of expanding: for every $x\in X$ the image of the ball $B(x,\delta_0)$ contains a neighborhood of the closure of $B(f(x),\delta_0)$.

Discussion on mathoverflow.

Answer 1

Following discussion on mathoverflow, this claim is false, but it seems that it is standard to make the extra assumption that $f(B(x, \delta_0))^\circ \supset \overline{B(f(x), \delta_0)}$ holds for all $x \in X$. After this, this seems to be quite easy.

First, we observe that $f(B(x, \delta_0))^\circ \supset \overline{B(f(x), \delta_0)}$ in particular implies the simpler $f(B(x, \delta_0)) \supset B(f(x), \delta_0)$. Together with the original expanding condition this implies $f(B(x, \delta)) \supset B(f(x), \delta)$ for all $\delta \leq \delta_0$. Namely, all elements of $B(f(x), \delta)$ will have a preimage in $B(x, \delta_0)$, and by the original expanding condition these cannot be at distance more than $\lambda^{-1}\delta$ from $x$, i.e. we even find preimages in $B(x, \lambda^{-1}\delta)$.

We also observe by a simple calculation that the extra assumption for $f$ implies it also for $f^n$, i.e. we in particular have $f^n(B(x, \delta)) \supset B(f^n(x), \delta)$ for all $\delta \leq \delta_0$ and $x \in X$.

Now take $U$ any open set. Take a smaller open set $U'$ inside $U$ so that $B(U', \delta) \subset U$ for some $\delta \leq \delta_0$. (For example, take a ball inside $U$ with radius $2\delta \leq \delta_0$, and use the corresponding $\delta$-ball.) Now cover the space $X$ with finitely many $\delta/2$-radius balls $V_i$, apply mixing finitely many times to get a transition time from $U'$ to any such ball, and take the maximum such time. We get that for some $n$, for any $V_i$ we can find $y \in U'$ such that $f^{n}(y) \in V_i$.

Now let $x \in X$ be arbitrary. Since the $V_i$ form a cover and have radius $\delta/2$, for any $x \in X$ we can find $V_i \subset B(x, \delta)$, so we can find $y \in U'$ such that $f^n(y) \in V_i \subset B(x, \delta)$.

Let $x' = f^n(y)$. We have $d(x', x) < \delta$. Now by the discussion at the beginning, and since $U$ contains $B(y, \delta)$, we have $f^n(U) \supset f^n(B(y, \delta)) \supset B(x', \delta) \ni x$, proving exactness since $x$ was arbirary.