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 $X$.

  • $\begingroup$ 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? $\endgroup$ Oct 4, 2021 at 2:12
  • $\begingroup$ 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. $\endgroup$ Oct 4, 2021 at 6:28
  • $\begingroup$ 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 $x$? $\endgroup$ Oct 4, 2021 at 13:08
  • $\begingroup$ 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. $\endgroup$ Oct 4, 2021 at 14:01

1 Answer 1


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.

  • $\begingroup$ Yes, the construction is very simple indeed! I guess one can insert the * symbol in any way as long as its upper density is zero. $\endgroup$ Oct 5, 2021 at 7:24

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