map $T$ on $X\backslash
S_0$ by setting $$T(x):=f_j(x)\quad \text{if $x\in I_j$}\,.$$ The map $T$ is left undefined on $S_0$. We also assume that

$$\label{2.1}
\big(\bigcup_{i\in\tA}J_i\big)\cap I_j=I_j\quad\forall j\,.$$

We introduce sets $X_j$, $S_j$, and $S$ by setting for $j\geq 1$ $$X_0:=[0,1]\,,\quad X_j:=X_{j-1}\backslash S_{j-1}\,,\quad
S_j:=\{x\in X_j\colon T(x)\in S_{j-1}\}\,,\quad S:=\bigcup_{j\geq
0}S_j\,.$$

\[lem2.1\] Under the condition , $T^n(X_{n+1})= X_1$ and $T(X\backslash S)=X\backslash S$. $X\backslash S$ is dense in $X$.

Condition is equivalent to $T(X_1)\supset X_1$. Since $X_2=X_1\backslash S_1$ and $S_1=\{x\in X_1\colon
T(x)\not\in X_1\}$, we have $T(X_2)=X_1$. Suppose that $T^n(X_{n+1})=X_1$; we prove that $T^{n+1}(X_{n+2})=X_{1}$. One has $X_{n+1}=X_{n+2}\cup S_{n+1}$ and $$X_1=T^n(X_{n+1})=T^n(X_{n+2})\cup T^n(S_{n+1})\,.$$ Applying once more $T$, $$X_1\subset T(X_1)=T^{n+1}(X_{n+2})\cup T^{n+1}(S_{n+1})\,.$$ $T^{n+1}$ is defined on $X_{n+1}$ and $S_{n+1}\subset X_{n+1}$. $$T^{n+1}S_{n+1}=\{x\in X_{n+1}\colon T^{n+1}(x)\in S_0\}= \{x\in
X_{n+1}\colon T^{n+1}(x)\not\in X_1\}\,.$$ Hence $T^{n+1}(X_{n+2})=X_{1}$. Clearly $T(X\backslash S)\subset
X\backslash S$ and $T(S\backslash S_0)\subset S$. Since $X_1$ is the disjoint union of $X\backslash S$ and $S\backslash S_0$, and $TX_1\supset X_1$, we have $T(X\backslash S)=X\backslash S$. The sets $X\backslash S_k$ are open and dense in $X$. By Baire’s Theorem $X\backslash S=\bigcap_{k}(X\backslash S_k)$ is dense.

Let $\Z_+:=\{0,1,2,\ldots\}$ and $\tA^{\Z_+}$ be equipped with the product topology. Elements of $\tA^{\Z_+}$ are called [strings]{} and denoted by $\ud{x}=(x_0,x_1,\ldots)$. A finite string $\ud{w}=(w_0,\cdots,w_{n-1})$, $w_j\in\tA$, is a [word]{}; we also use the 