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I am trying to understand the concept of topological entropy as defined by $h_{top} (f) = \lim_{\epsilon \rightarrow 0^+} \lim_{n \rightarrow \infty} \frac{1}{n} \log(sep(n, \epsilon, f))$, where $sep(n,\epsilon,f)$ is the maximum cardinality of an $(n,\epsilon)$-spanning set, which can be replaced with minimum spanning set cardinality or the minimum covering set cardinality. Especially the $\epsilon$ limit causes some problems when i try to actually determine the entropy of a map.

For instance, the circle rotations $R_{\alpha}:S^1 \rightarrow S^1$, defined as $R_{\alpha}(x) = x+\alpha \mod 1$, should have entropy 0. However, if I am correct, the set $\{\epsilon, 2\epsilon, \ldots, 1\}$ is an $(n,\epsilon)$-seperated set with cardinality $\frac{1}{\epsilon}$ when $\epsilon = 1/c$ for some $c \in \mathbb{R}$. Since the circle rotation does not change the distance between points, $sep(n,\epsilon,f) = sep(1,\epsilon,f)$. But now the entropy is $$ h(R_{\alpha}) = \lim_{\epsilon \rightarrow 0^+} \lim_{n \rightarrow \infty} \frac{1}{n} \log(sep(1, \epsilon, f)) \geq \lim_{\epsilon \rightarrow 0^+} \lim_{n \rightarrow \infty} \frac{-\log(\epsilon)}{n}. $$ I do not see how to show that the entropy should be 0. In fact, I do not even see or understand why it is true in this case. How should i continue? Find a different spanning/covering/seperating set?

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    $\begingroup$ If you want a upper bound, use the "minimum cardinality of a covering set" definition. $\endgroup$ – D. Thomine Jan 12 at 15:14
  • $\begingroup$ Then inserting $2\epsilon$ for the "minimum cardinality of a covering set" would give me an upper bound, right? The limit stays the same, so does that mean the limit in my original question is in fact 0? I did not think that I could simply let $\lim_{\epsilon\rightarrow 0^+}\lim_{n\rightarrow\infty} \frac{-\log(\epsilon)}{n} = 0$. $\endgroup$ – user304122 Jan 12 at 15:42
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    $\begingroup$ Yes, the limit would be 0. $\endgroup$ – D. Thomine Jan 12 at 17:34
  • $\begingroup$ Thank you very much! Question answered. $\endgroup$ – user304122 Jan 12 at 17:42

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