# Non-minimal system in which every point is a full entropy point

Is there a discrete topological dynamical system $$(X,f)$$, where $$X$$ is a compact metric space (with distance $$d$$), which is transitive but not minimal, such that $$h(f)>0$$ and every point is a full entropy point?

By transitive I mean that there is a point with a dense orbit. By minimal I mean that every orbit is dense. By full entropy point I mean the concept defined in 1, that is a point $$x$$ such that the topological entropy, restricted to any of its closed neighborhoods, coincides with the entropy $$h(f)$$ of the system.

Finally, for every closed neighbor $$K(x)$$ of $$x$$, let $$N(\epsilon, n, K)$$ be the largest cardinality of an $$(n,\epsilon)$$-separated subset $$A\subset K$$ in the metric $$d_n(x,y)=\max_{i=0,\dots,n}d(f^i(x),f^i(y))$$. Then the entropy restricted to $$K(x)$$, in symbols $$h(f,K)$$, is defined as the usual limit $$\lim_{\epsilon\to 0}\lim_{n\to\infty} \frac 1n\log{N(\epsilon, n, K)}.$$ Notice that $$h(f)=\sup_K h(f,K)$$, where the supremum is taken over all compact subsets of $$X$$.

• To get some clarification: what properties of f do you want? If $f$ is continuous and a self-map of $[0,1]$, then I think a result of Vellekoop and Berglund shows that if $f$ is transitive, then $f$ has dense periodic points. In particular, $f$ is automatically not minimal, and has points which are periodic. I'm not sure I understand your definition of full entropy point (I don't see how to restrict to a neighborhood $K(x)$ which is generally not invariant), but it feels like periodic points should have $0$ entropy, right? Oct 4, 2021 at 2:12
• Thanks, I added further clarifications. In the definition I meant for entropy restricted to $K(x)$, taken from a paper by X.Ye and G.Zhang which I linked, it is not required that $K(x)$ is invariant. I also replaced the unit interval by an arbitrary compact metric space since the classical result by Vellekoop that you mentioned answered in fact in the negative. Oct 4, 2021 at 6:28
• Is it really true that periodic points have zero/lower entropy for your definition? I'm now not so sure now that I've seen the definition you're using, so it's not clear that Vellekoop-Berglund actually negatively answer the original question. I mostly ask because I think I see an easy subshift example for your original question, but it may be due to a lack of understanding of your $K(x)$ entropy. Is it true that if $X$ is a subshift, then the "entropy at $x$" just comes from the exponential growth rate of numbers of subwords of $x$? Oct 4, 2021 at 13:08
• Yes, I would think so, it depends on the asymptotic exponential growth rate of the number of subwords of $X$. In any case Vellekoop-Berglund shows that the interval case less rich, so I believe the current formulation of the question is more interesting. Oct 4, 2021 at 14:01

I guess this is not hard to do with a subshift. Take $$X$$ to be your favorite minimal positive entropy subshift on $$\{0,1\}$$ (such examples are constructed in Hahn-Katznelson, among other works).
Now, choose any $$x \in X$$ and define $$y$$ on $$\{0,1,*\}$$ as $$y = .* \ x_1 * x_2 x_3 * x_4 x_5 x_6 * x_7 x_8 x_9 x_{10} * \ldots$$
Define $$Y$$ to be the orbit closure of $$y$$. Clearly $$Y$$ is transitive by definition, and is not minimal since it strictly contains $$X$$. It's not hard to check that $$h(Y) = h(X)$$; by the ergodic theorem, any ergodic measure on $$Y$$ gives $$*$$ zero measure, and so is supported on $$X$$. Then $$h(Y) = h(X)$$ by the variational principle.
Finally, every point of $$Y$$ should have full entropy, since every point $$z \in Y$$ contains arbitrarily long words of $$X$$, and so all words in the language of $$X$$ by minimality. So if I understand your definition correctly, $$z$$ should have full entropy.
I don't see how any of this could be carried over to $$[0,1]$$, but as we discussed in the comments, it may be impossible to construct such an example on the interval.