A systems $<f_0,f_1>$ is minimal if the set $\{h(x): h=f_{i_n}\circ f_{i_{n-1}}\circ...\circ f_{i_1}, i_k \in \{0,1\},n>0\}$ is dense in $S^1$, for every $x\in S^1$. Consider $f:S^1 \to S^1, f(x)=2x (mod \, 1)$ and $f_{\epsilon}=f+ \epsilon$. Is there an $\epsilon>0$ such that $<f,f_{\epsilon}>$ is minimal?
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$\begingroup$ Thanks by your answer. But I have another question: If $f_{\epsilon}$ is ${\epsilon} \ \ C^1$-closed from $f$ with $f(x)\not=f_{\epsilon}(x)$ for all $x \in S^1$, then $<f,f_{\epsilon}>$ is minimal? $\endgroup$– Wagner RanterMay 29, 2016 at 16:38
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$\begingroup$ No, if I understand what you are asking. I assume you want to know whether the conclusion is true when $\epsilon$ is small enough, but if you let $f_\epsilon = f + q$ for some nonzero $q \in \mathbb{Q}$, then the $\langle f, f_\epsilon \rangle$-orbit of $0$ is finite. $\endgroup$– Dave Witte MorrisMay 30, 2016 at 2:26
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Yes, for example, you could choose $\epsilon$ so that $\{2^k \epsilon \mid k = 1,2,3,\ldots\}$ is dense in $S^1$. By letting all but one of the $f_{i_k}$'s be $f$, we see that the orbit of $x$ under words of length $n$ contains $f^n(x) + 2^k\epsilon$ for $1 \le k \le n$.