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I am reffering to this site: http://www.scholarpedia.org/article/Topological_entropy

Definitionj of topological Entropy by Adler, Kohnheim

For an open cover $\mathcal{U}$ of $X$, let $N(\mathcal{U})$ denote the smallest cardinality of a subcover of $\mathcal{U}$. If $\mathcal{U}$ and $\mathcal{V}$ are open covers of $X$, then $$ \mathcal{U}\vee\mathcal{V}=\left\{U\cap V: U\in\mathcal{U},V\in\mathcal{V}\right\} $$ is called their common refinement.

Let $$ \mathcal{U}^n=\mathcal{U}\vee T^{-1}\mathcal{U}\vee T^{-2}\mathcal{U}\vee\cdots\vee T^{-(n-1)}\mathcal{U}, $$ where $T^{-k}\mathcal{U}=\left\{T^{-k}U: U\in\mathcal{U}\right\}$.

Define the entropy of $T$ relative to $\mathcal{U}$ as $$ h(\mathcal{U},T)=\lim_{n\to\infty}\log\frac{1}{n}N(\mathcal{U}^n). $$ The topological entropy of $T$ is defined as $$ h(T)=\sup_{\mathcal{U}}h(\mathcal{U},T), $$ where the supremum ranges over all open covers $\mathcal{U}$ of $X$.

In order to understand this definitions, I looked at the left sided shift, i.e. let $(X,T)$ be a subshift, i.e., T being the left shift map on a closed left shift invariant collection $X$ of sequences of symbols belonging to a finite alphabet.

It is said that the topological entropy of $T$ is given by $$ h(t)=\lim_{n\to\infty}\frac{1}{n}\log \#\mathcal{B}_n, $$ where $\mathcal{B}_n$ denotes the collection of all words of length $n$ appearing anywhere in the sequences $x\in X$.

Could you please explain me how I get from the definition of topological entropy (as given above) that $$ h(t)=\sup_{\mathcal{U}}h(\mathcal{U},T)=\lim_{n\to\infty}\frac{1}{n}\log \#\mathcal{B}_n? $$

  • You need a result saying that if $\mathcal{U_i}$ is a sequence of open cover such the maximal diameter of the elements in $\mathcal{U_i}$ tends to $0$ as $i$ tends to infinity, then $h(T)=lim_{i\rightarrow\infty}h(\mathcal{U_i},T)$. Then you can make such covers using cylinder sets. – Siming Tu Apr 15 '15 at 16:29
  • There is such a statement in Walters: Let $(X,d)$ be a compact metric space. If $\left{\alpha_n\right}_{n=1}^{\infty}$ is a sequence of open covers of $X$ with $\text{diam}(\alpha_n)\to 0$ then if $h(T)<\infty$ $\lim_n h(T,\alpha_n)$ exists and equals $h(T)$. - - - But what is the metric here and how to choose $\left{\alpha_n\right}$ in order to show my result? –  Apr 15 '15 at 16:40
  • The metric on the space is that for two sequence of symbols $x=(x_i)$ and $y=(y_i)$, $d(x,y)=0$ when $x=y$ and $d(x,y)=1/i$ when $x\neq y$ and $i=min{j:x_j\neq y_j}$. – Siming Tu Apr 15 '15 at 17:07
  • For the partition you can make partition into cylinders which in each cylinder they have the same first $n$ coordinates. – Siming Tu Apr 15 '15 at 17:12
  • If you know what a generator is, you take a look at the book of Peter Walters(from your comment I see that you have one copy), in the Chapter of topological entropy he gives a proof using the notion of generator, which is discussed also in that chapter. You can take a look and I think you can understand it after reading that. – Siming Tu Apr 15 '15 at 17:18
  • Using your first idea.. it then is $h(T)=\lim_{n}h(T,\alpha_n)$. Now I think that $h(T,\alpha_n)=\lim_n \frac{1}{n}\log# \mathcal{B}_n$. - -What then happend with the second $\lim_n$? –  Apr 15 '15 at 17:19
  • In fact that $h(T,\alpha_n)$ are all equal. – Siming Tu Apr 15 '15 at 17:23
  • Ok. Another question. What if $X=\left{0,1,2\right}^{\mathbb{Z}}$ and T acting like this: a 0 becomes a 1 if at least one of its two neighours is 1, a 1 becomes a 2 and a 2 becomes a 0. What then $h(T)$? Same trick? But whats $h(T,\alpha_n)$ here? –  Apr 15 '15 at 17:26
  • Let us continue this discussion in chat. – Siming Tu Apr 15 '15 at 17:27

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