## Appendix A Proofs of Theorems and Corollaries
Then, \(d(s,t)\leq\frac{1}{2^{n}}\). Conversely, if \(d(s,t)<\frac{1}{2^{n}}\), then \(s_{i}=t_{i}\) for \(i=0,1,\ldots,n\)._  
**Proposition A3** ([PERSON] [15]). : \(\sigma:\Sigma_{2}\to\Sigma_{2}\) _is continuous._  
**Definition A3** ([PERSON] [15]). : _The itinerary of \(x\) is a sequence \(S(x)=s_{0}s_{1}s_{2}\ldots\) where \(s_{j}=0\) if \(f^{j}(x)\in I_{0}\), \(s_{j}=1\), if, \(f^{j}(x)\in I_{1}\)._  
Thus, the itinerary of \(x\) is an infinite sequence of \(0\)s and \(1\)s. That is, \(S(x)\) is a point in the sequence space \(\Sigma_{2}\). We think of \(S\) as a map from \(\Lambda\) to \(\Sigma_{2}\). This map has several interesting properties. Let \(J\subseteq I_{0}\) or \(J\subseteq I_{1}\) be a closed interval. Let \(f^{-n}(J)=\{x\in I_{0}\cup I_{1}|f^{n}(x)\in J,n\geq 1\}\). \(f^{0}(J)=J\). Since item \(A1\) above holds, \(f^{-n}(J)\) exists as the preimage of \(J\) under functions \(f^{n}(x)\). Let \(I_{s_{0}s_{1}\ldots s_{n}}=\cap_{0<k\leq n}f^{-k}(I_{s_{k}})\)  
\(=\cap_{0<k\leq n}(f^{k}(I_{0}\cup I_{1})\cap I_{s_{k}})=\{x\in I_{0}\cup I_{1 }|x\in I_{s_{0}}f(x)\in I_{s_{1}},f^{2}(x)\in I_{s_{2}},\ldots,f^{n}(x)\in I_{s _{n}}\}=I_{s_{0}}\cap f^{-1}(I_{s_{1}})\cap\ldots\cap f^{-n}(I_{s_{n}})=I_{s_{0 }}\cap(f(I_{0}\cup I_{1})\cap I_{s_{1}})\cap\ldots\cap(f^{n}(I_{0}\cup I_{1}) \cap I_{s_{n}})\). \(f^{-1}(J)\) consists of two closed subintervals, one in \(I_{0}\) and one in \(I_{1}\). Hence, \(f^{-1}(I_{0}\cup I_{1})\) consists of four disjoint closed subintervals, two in \(I_{0}\) and two in \(I_{1}\). Similarly, in general, \(f^{-n}(J)\) consists of \(2^{n}\) closed subintervals, \(2^{n-1}\) in \(I_{0}\) and \(2^{n-1}\) in \(I_{1}\). Hence, \(f^{-n}(I_{0}\cup I_{1})\) consists of \(2^{n+1}\) disjoint closed subintervals, \(2^{n}\) in \(I_{0}\) and \(2^{n}\) in \(I_{1}\). Then, the sequence \(I_{s_{0}s_{1}\ldots s_{n}}\) is not empty. We have \(\cap_{0\leq k\leq n}f^{-k}(I_{0}\cup I_{1})=\cap_{0\leq k\leq n}(f^{k}(I_{0} \cup I_{1})\cap(I_{0}\cup I_{1}))=\cap_{0\leq k\leq n}f^{k}(I_{0}\cup I_{1})= \cup_{s_{0}s_{1}\ldots s_{n}}\). First, we consider the properties of the sequence \(I_{s_{0}s_{1}\ldots s_{n}}=I_{s_{0}}\cap f^{-1}(I_{s_{1}s_{2}\ldots s_{n}})\) and \(I_{s_{0}s_{1}\ldots s_{n}}=I_{s_{0}s_{1}\ldots s_{n-1}}\cap f^{-n}(I_{s_{n}}) \subset I_{s_{0}s_{1}\ldots s_{n-1}}\) hold. We may assume that \(I_{s_{1}s_{2}\ldots s_{n}}\) is a nonempty subinterval so that, by the observation above, \(f^{-1}(I_{s_{1}s_{2}\ldots s_{n}})\) consists of two closed intervals, one in \(I_{0}\) and one in \(I_{1}\). Hence, \(I_{s_{0}}\cap f^{-1}(I_{s_{1}s_{2}\ldots s_{n}})\) is a single closed interval. \(I_{s_{0}s_{1}\ldots s_{n}}\) forms a nested sequence of nonempty closed intervals because \(I_{s_{0}s_{1}\ldots s_{n}}=I_{s_{0}s_{1}\ldots s_{n-1}}\cap f^{-1}(I_{s_{n}}) \subset I_{s_{0}s_{1}\ldots s_{n-1}}\). Therefore, we conclude that \(\cap_{0\leq n\leq\infty}I_{s_{0}s_{1}\ldots s_{n}}\) is nonempty closed, set and \(\Lambda\) is also nonempty closed set such that \(\Lambda=\cap_{0\leq n\leq\infty}f^{n}(I_{0}\cup I_{1})=\cap_{0\leq n\leq \infty}f^{-n}(I_{0}\cup I_{1})=\cup_{s_{0}s_{1}\ldots s_{n},0}(\cap_{0\leq n\leq \infty}I_{s_{0}s_{1}\ldots s_{n}})\). Secondly, we show that \(S\) is one to one. Let \(x,y\in\Lambda\) and suppose \(S(x)=S(y)\). Then, for each \(n\), \(f^{n}(x)\) and \(f^{n}(y)\) lie on the same interval \(I_{0}\) or \(I_{1}\). This implies that \(f\) is monotone on the interval between \(f^{n}(x)\) and \(f^{n}(y)\). Consequently, all points in this interval remain in \(I_{0}\cup I_{1}\) when we apply \(f\). Now, \(|Df|>K>1\) at all points in this interval; thus, as in item \(A2\), each iteration of \(f\) expands this interval by a factor of \(K\). Hence, the distance between \(f^{n}(x)\) and \(f^{n}(y)\) grows without bound, so these two points \(f^{n}(x)\) and \(f^{n}(y)\) must eventually lie on \(I_{0}\) and \(I_{1}\) (or \(I_{1}\) and \(I_{0}\)), respectively. This contradicts the fact that they have the same itinerary. Thirdly, to see that \(S\) is onto, let \(s=(s_{0}s_{1}s_{1}\ldots)\).