Let $f:X \to X$ be a continuous map, where $X$ is a compact metric space. We say that $f$ is (*locally*) **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 some lecture notes that I have and searching the internet, it seems that for an expanding map the definitions of (topological) exactness and (topological) mixing are equivalent.

- (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 $n \ge n_0$.
- (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 (11.2.2 from Foundations of Ergodic Theory by Viana and Oliveira) 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... My main strategy was to try to build a sequence that has no convergent subsequence, but I couldn't.

For clarification, $f^n(x)=f\circ f \circ \cdots \circ f(x)$ an iterative map. I also posted the question on math.stackexchange but didn't get any hints.

**Edit:** I found a hit for the exercise in Viana and Oliveira's book. But an additional assumption is required 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 $B(f(x),\delta_0)$.

Show that if $f : X \to X$ is topologically mixing then the set of periodic points is dense. This allows using Theorem 11.2.15. As $f$ is topologically mixing and $M_1$ is open invariant, we have that $M_1 = M$, that is $k=1$. Similarly, show that for $f$ to be topologically mixing it is necessary that $m(1) = 1$. Then, by Theorem 11.2.15, the transformation $f : X \to X$ is topologically exact.