A dynamical system is proximal if $$\:\forall (x,y) \in X \times X, \: \liminf_{n \rightarrow \infty} d(f^{n}(x),f^{n}(y)) = 0 $$ (where $X$ is a compact metric space with metric $d$). Is it true that the topological entropy $h(f)$ is $0$? If not, what is a counterexample?
1 Answer
Maybe there is an easier example, but here is an example of a proximal system with positive entropy. The dynamical system is a socalled subshift $(X, \sigma)$, where $X$ is a closed shiftinvariant subset of $\{0,1,2\}^{\mathbb{N}}$ and $\sigma$ is the left shift map sending any sequence $x = .x_1 x_2 x_3 \ldots$ to $\sigma x = .x_2 x_3 \ldots$. The product topology is induced by the metric $d(x,y) := \sum \frac{x_i  y_i}{2^i}$.
Define a sequence of finite words as follows: $w_1 = 1$, and for all $n$, $w_{n+1} = w_n 0^n w_n$. So $w_2 = 101$, $w_3 = 10100101$, etc. Then $w_n$ approach a limit sequence $x = 1010010100010100101 \ldots$. Define $X = \overline{\sigma^n x}$, the orbit closure of $x$. Then $(X, \sigma)$ is a subshift.
Denote by $L_n$ the length of $w_n$; it's a routine induction that $L_n = 1.5 \cdot 2^n  n  1$. Also, it's easy to see that $w_n$ has $2^{n1}$ $1$ symbols.
It's again a fairly short induction to see that for any $n$, $x$ is a concatenation of $w_n$ and runs of $0$s of length at least $n$. Therefore, any subword of $x$ of length $2n + L_n$ contains $0^n$.
Now, consider any $y,z \in X$. The first $2(2n + L_n) + L_{2n + L_n}$ symbols of $y$ are a subword of $x$ since $X = \overline{\sigma^n x}$. Therefore, they contain a substring of $0^{2n + L_n}$ zero symbols. But then the subword of $z$ occupying those locations is of length $2n + L_n$, and was a subword of $x$, so it contains a string of $n$ zero symbols. Therefore, there is a location $k$ after which $y$ and $z$ both contain $n$ consecutive $0$s, meaning that $\sigma^k y$ and $\sigma^k z$ agree on the first $n$ symbols, so $d(\sigma^k y, \sigma^k z) \leq \sum_{n+1}^{\infty} \frac{2}{2^i} = 2^{n+1}$. Since $n$ was arbitrary, $\liminf d(\sigma^k y, \sigma^k z) = 0$, and since $y,z$ were arbitrary, $(X, \sigma)$ is proximal.
Now, $(X, \sigma)$ actually has zero entropy, but we will make a simple modification to it to introduce entropy. Define $X' \subset \{0,1,2\}^\mathbb{N}$ to be the set of all sequences which map to a sequence in $X$ when all $2$s are changed to $1$s. Alternately, $X'$ is obtained by taking all sequences in $X$, and letting some $1$s change to $2$s.
Since all $0$ locations in sequences in $X$ were unchanged in this operation, nothing about the above argument changes, and $(X', \sigma)$ is still proximal. However, it has positive entropy.
Since $(X', \sigma)$ is a subshift, one can compute topological entropy by counting legal words. If $N_k$ is the number of words appearing in points of $X'$ of length $k$, then $h(X) = \lim_k \frac{\log N_k}{k}$. Notice that $w_n$ was a legal word in $X$ with $2^{n1}$ $1$ symbols and length $L_n = 1.5 \cdot 2^n  n  1$. This word gives rise to $2^{2^{n1}}$ legal words in $X'$ (by changing any subset of $1$s to $2$s), and so $N_{L_n} \geq 2^{2^{n1}}$. But then
$h(X') = \lim_k \frac{\log N_k}{k} = \lim_n \frac{\log N_{L_n}}{L_n} \geq \limsup_n \frac{\log 2^{2^{n1}}}{1.5 \cdot 2^n} = \frac{\log 2}{3} > 0$.


$\begingroup$ No problem! If the answer sounds right, can you accept it, so others know the question has been answered? $\endgroup$ Nov 20, 2023 at 21:52