Topological entropy of circle homeomorphism is zero. True or false?

Question

may I know if it is true that $\ f: S^1 \to S^1$ a homeomorphism, then $h_{top}(f) = 0$, where $h_{top}$ stands for topological entropy.

I believe this statement is true, but I cannot prove it.

Answer 1

Here, we use the spanning set definition of topological entropy. Since $f$ is uniformly continuous on $S^1$, for arbitrary $\epsilon>0$, there exists a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ if $|x-y|\leq\delta$. Let $S_n$ be the cardinality of the minimal $(n,\epsilon)-spanning$ set. Then we study the relations between $S_n$ and $S_{n+1}$.

Denote by $\{I_i,i=1,...,S_n\}$ the open interval cover of $S^1$ corresponding to the minimal $(n,\epsilon)-spanning$ set. We consider an open interval $I_i=(x,y)$. Then we have $d(f^i(x),f^i(y))<\epsilon, i=0,1,...,n.$ Therefore, it's critical to determine $d(f^{n+1}(x),f^{n+1}(y))<\epsilon$. In order to ensure $d(f^{n+1}(x),f^{n+1}(y))<\epsilon$, we devide the interval $(f^{n}(x),f^{n}(y))$ or $(f^{n}(y),f^{n}(x))$ into $\epsilon/\delta$ subintervals. Then each of such subintervals has a preimage interval of $f^n$ in $I_i=(x,y)$ for $f$ is a homeomorphism. Thus we obtain $S_n\cdot \epsilon/\delta$ intervals cover $S^1$ such that $d(f^i(x),f^i(y))<\epsilon, i=0,1,...,n+1,$ for $x,y$ in the same interval. By the minimality of spanning set, so we have $S_{n+1}\leq S_n\cdot \epsilon/\delta.$ By the definiton of topological entropy $h(f)$, $$0\leq h(f)=\lim_{\epsilon\rightarrow 0}\overline{\lim_{n\rightarrow\infty}}\frac{\log{S_n}}{n}\leq\lim_{\epsilon\rightarrow 0}\epsilon/\delta=0.$$ Hence $h(f)=0.$