that there exist $x\in X\backslash S$ and $k$ such that $\sigma^k\i(x)= \ud{v}^{\i_k(x)}$. Since $(\sigma^k\i(x))_0=\i_0(T^kx)$, we can assume, without restricting the generality, that $k=0$ and $\i_0(x)=j$. Therefore $x\in (a_{j}, a_{j+1})$, and for all $y\in X\backslash S$, such that $x\leq y<a_{j+1}$, we have by Lemma \[lem2.4\] that $\i(y)=\i(x)=\ud{v}^{j}$. By Theorem \[thm2.1\] this contradicts the hypothesis that the $\varphi$-expansion is valid. The other case, $\sigma^k\i(x)= \ud{u}^{\i_k(x)}$, is treated similarly. This proves half of the first statement. The second half is a consequence of Proposition \[prof2.2\]. The second statement also follows, as well as the third, since $T(X\backslash S)=X\backslash S$ (we assume that holds).

Let $\ud{x}\in\tA^{\Z_+}\backslash\Sigma$ and $m\in\Z_+$ be the smallest integer such that one of the conditions defining $\Sigma$ is not verified. Then either $\sigma^m\ud{x}\preceq\underline{u}^{x_m}$, or $\sigma^m\ud{x}\succeq \underline{v}^{x_m}$. The map $\overline{\varphi}_\infty$ is continuous (Theorem \[thm2.1ter\]). Hence, for any $j\in\tA$, $$\overline{\varphi}_\infty(\ud{u}^j)=a_{j}\quad\text{and}\quad
\overline{\varphi}_\infty(\ud{v}^j)=a_{j+1}\,.$$ Let $\sigma^m\ud{x}\preceq\underline{u}^{x_m}$. Since $\ud{v}^{x_m-1}\prec\sigma^m\ud{x}$, $$a_{x_m}=\overline{\varphi}_\infty(\ud{v}^{x_m-1})
\leq\overline{\varphi}_\infty(\sigma^m\ud{x})\leq
\overline{\varphi}_\infty(\ud{u}^{x_m})=a_{x_m}\,.$$ The other case is treated in the same way. From definition $\ud{u}^{u^j_n}\preceq \sigma^n\ud{u}^j\preceq
\ud{v}^{u^j_n}$. Suppose that $\delta(\ud{u}^j_0\cdots\ud{u}^j_{n-1})=1$ and $\sigma^n\ud{u}^j=\ud{v}^{u^j_n}$. By continuity of the $\varphi$-code there exists $x\in X\backslash S$ such that $x>a_j$ and $\i_k(x)=\ud{u}^j_k$, $k=0,\ldots,n$. Let $a_j<y<x$. Since $\delta(\ud{u}^j_0\cdots\ud{u}^j_{n-1})=1$, $T^ny<T^nx$ and consequently $$\lim_{y\downarrow
a_j}\i(T^ny)=\sigma^n\ud{u}^j\preceq\i(T^nx)\preceq
\ud{v}^{u^j_n}\,.$$ Hence $\sigma^n\i(x)= \ud{v}^{x_n}$, which is a contradiction. The other cases 