# Equivalence of the definitions of exactness and mixing

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.

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 (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.

• After the extra assumption it seems to be straightforward, I wrote an answer on stackexchange. Commented May 4, 2023 at 4:33

edit As the author clarifies in the comments below, there is indeed a hidden assumption. In the case of a subshift, as Vaughn Climenhaga explains the extra condition means exactly that the subshift is finite type. For completeness I'll give the trivial proof that exact = mixing in this case. Perhaps it inspires the poster to find the general proof.

Theorem. Mixing and exactness are equivalent for one-sided subshifts of finite type.

Proof. As explained by Vaughn, the extra condition defining subshifts of finite type amounts to $$\sigma([w]) = [\sigma(w)]$$ for all words of length at least $$n$$. Writing this condition as "$$a[w] = [aw]$$ for $$a$$ in the alphabet, and $$w$$ of length at least $$n$$", we see by induction that the same holds for also longer words in place of $$a$$, i.e. $$u[w] = [uw]$$ for all $$u, w$$ (when $$uw$$ is in the language of the subshift and $$w$$ is long enough).

It suffices to prove that mixing implies exact. Let $$U$$ be any open set. Using mixing, for each word $$[w_i]$$ of length $$n$$ find $$n_{0,i}$$ such that $$\sigma^n(U) \cap [w_i] \neq \emptyset$$ for $$n \geq n_{0,i}$$. Take $$n_0 = \max(n_{0,i})$$. Then for any $$x \in X$$, you can find some $$y \in X$$ such that $$y \in U$$, and $$\sigma^{n_0}(y) \in [w]$$ where $$w$$ is the length-$$n$$ prefix of $$x$$.

Now in the situation of the previous paragraph we have $$[uw] = u[w]$$ where $$u$$ is the prefix of $$y$$ up to $$w$$. In particular, since $$x \in [w]$$ we have $$y = ux \in u[w] = [uw] \subset X$$ so $$y \in X$$ satisfies $$\sigma^{n_0}(y) = x$$. Since $$x$$ was arbitrary, $$\sigma^{n_0}(U) = X$$ proving exactness. Square.

original I think the claim is simply false. Are there hidden assumptions?

Consider the shift map on a subshift (closed shift-invariant set) $$X \subset A^{\mathbb{N}}$$ where $$A$$ is a finite set, considered with define the metric defined by $$d(x, y) = 2^{-i}$$ when $$i$$ is maximal such that $$x_{[0,i)} = y_{[0,i)}$$, and with the shift map $$\sigma(x)_i = x_{i+1}$$. Taking $$\lambda = 2$$ and $$\delta_0 = 1/2$$ this is expanding.

Now mixing for a subshift $$X$$, expressed in terms of the usual cylinder basis, amounts to saying that for any two words $$u, v$$, there exists $$n_0$$ such that we can find for each $$n \geq n_0$$ a point $$x \in X$$ such that $$x = uwvy$$ for some word $$w$$ of length $$n$$, and some tail $$y \in X$$. Exact means that for each word $$u$$ there exists $$n_0$$ such that we can find for each $$y \in X$$ a point $$x \in X$$ such that $$x = uwy$$ for some word $$w$$ of length $$n_0$$.

Let $$A = \{0,1,2\}$$ and take $$X$$ to be the subshift defined by the rule that whenever $$v10^n$$ appears in an element of $$X$$ with $$v$$ of length $$n$$ (as a finite subword, in consecutive positions), we must have $$v = 0^n$$. We can think that each $$1$$-symbol counts the number of $$0$$s on its right, and forces the same number of $$0$$s on its left.

This is mixing: if $$u, v$$ are any two words appearing in $$X$$, and $$v$$ has length $$k$$, then $$u2^m0^kv1111... \in X$$ for any positive $$m$$. Every $$1$$-symbol is happy: there cannot be a problem with those in $$v$$ because they see either internal $$0$$-runs, or the $$0^k$$ we put there. The $$1$$s in $$u$$ see only internal $$0$$s so again there is no problem.

On the other hand it's not exact. Take $$u = 2$$. You actually cannot stick the tail $$y = 100000...$$ after it with any choice of $$n_0$$.

• I don't believe it has any hidden assumptions. I first came across this issue, when I saw it in an article, instead of the usual definition of topologically mixing was the definition of topologically exact. So I found this exercise in Viana's book. Commented May 2, 2023 at 4:09
• Ok, then you can peruse the counterexample, ask me if something is wrong/unclear. I was specifically wondering if compact metric should include something like connected, not that I immediately know whether that changes the answer. Commented May 2, 2023 at 4:11
• I confess that I did not understand your example very well. Looking at Viana's book, I noticed that he requires an additional hypothesis to be expansive: for every $x \in X$ the image of the ball $B(x,\delta_0)$ contains a neighborhood of the closure of $B(f(x),\delta_0)$. Does this compromise your counterexample? Commented May 3, 2023 at 16:08
• A less pedantic comment: I don't believe the counterexample in this answer has the extra property that Viana and Oliveira demand of "expanding maps". For a shift space $X$, balls become cylinders, and for every finite word $w$ we have $\sigma[w] \subset [\sigma w]$, where $\sigma w$ denotes the word $w$ with the first symbol removed. In general the inclusion may be proper, in which case the word $w$ is a left constraint, to use the terminology of Buzzi's 2005 Inventiones paper (see link.springer.com/article/10.1007/s00222-004-0392-1 or arxiv.org/abs/math/0305164). .......... Commented May 4, 2023 at 1:09
• But this example does certainly illustrate that without this extra property, maps that merely expand distances do not necessarily behave the way that we might expect if we think first of expanding maps on manifolds. Personally I think it is better to make this extra property explicit rather than wrapping it into the word "expanding", which taken at face value would not seem to include such a condition, but it is quite common in the literature to use the word "expanding" in this way. Commented May 4, 2023 at 1:13