Topological entropy of a circle rotation

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?

• If you want a upper bound, use the "minimum cardinality of a covering set" definition. – D. Thomine Jan 12 at 15:14
• 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 $\lim_{\epsilon\rightarrow 0^+}\lim_{n\rightarrow\infty} \frac{-\log(\epsilon)}{n} = 0$. – user304122 Jan 12 at 15:42
• Yes, the limit would be 0. – D. Thomine Jan 12 at 17:34
• Thank you very much! Question answered. – user304122 Jan 12 at 17:42