# Does every proximal dynamical system have zero topological entropy?

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?

Maybe there is an easier example, but here is an example of a proximal system with positive entropy. The dynamical system is a so-called subshift $$(X, \sigma)$$, where $$X$$ is a closed shift-invariant 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 $$2^{n-1}$$ $$1$$ 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 $$2^{n-1}$$ $$1$$ symbols and length $$L_n = 1.5 \cdot 2^n - n - 1$$. This word gives rise to $$2^{2^{n-1}}$$ legal words in $$X'$$ (by changing any subset of $$1$$s to $$2$$s), and so $$N_{L_n} \geq 2^{2^{n-1}}$$. But then

$$h(X') = \lim_k \frac{\log N_k}{k} = \lim_n \frac{\log N_{L_n}}{L_n} \geq \limsup_n \frac{\log 2^{2^{n-1}}}{1.5 \cdot 2^n} = \frac{\log 2}{3} > 0$$.

• Thanks, that rly help Nov 20, 2023 at 20:40
• No problem! If the answer sounds right, can you accept it, so others know the question has been answered? Nov 20, 2023 at 21:52