conclude that $\ud{x}^\prime\preceq\i(x)$. If $y_\downarrow(\ud{x}^\prime)$ is not a cluster point, then we use the cluster point $y_\uparrow(\ud{x}^\prime)<y_\downarrow(\ud{x}^\prime)$ for the argument.

\[thm2.1\][[@P2]]{} A $\varphi$-expansion is valid if and only if the $\varphi$-code $\i$ is injective on $X\backslash S$.

Suppose that the $\varphi$-expansion is valid. If $x\not=z$, then $$x=\overline{\varphi}\big(\i_0(x)+
\overline{\varphi}(\i_1(x)+\ldots)\big)\not=
\overline{\varphi}\big(\i_0(z)+
\overline{\varphi}(\i_1(z)+\ldots)\big)=z\,,$$ and therefore $\i(x)\not=\i(z)$. Conversely, assume that $x\not=z$ implies $\i(x)\not=\i(z)$. Let $x\in X\backslash S$, $\ud{x}=\i(x)$, and suppose for example that $y_\uparrow(\ud{x})<
y_\downarrow(\ud{x})$ are two cluster points. Then by Lemma \[lem2.4quatro\] any $y$ such that $y_\uparrow(\ud{x})<y<
y_\downarrow(\ud{x})$ is in $X\backslash S$ and $\i(y)=\ud{x}$, contradicting the hypothesis. Therefore $z:=\lim_n\overline{\varphi}_n(\ud{x})$ exists. If $z\not=x$, then we get again a contradiction using Lemma \[lem2.4quatro\].

Theorem \[thm2.1\] states that the validity of the $\varphi$-expansion is equivalent to the injectivity of the map $\i$ defined on $X\backslash S$. One can also state that the validity of the $\varphi$-expansion is equivalent to the surjectivity of the map $\overline{\varphi}_\infty$.

\[thm2.1bis\] A $\varphi$-expansion is valid if and only if $\overline{\varphi}_\infty\colon \tA^{\Z_+}\ra [0,1]$ is well-defined on $\tA^{\Z_+}$ and surjective.

Suppose that the $\varphi$-expansion is valid. Let $\ud{x}\in\tA^{\Z_+}$ and suppose that $\{\overline{\varphi}_n(\ud{x})\}_n$ has two different accumulation points $y_\uparrow<y_\downarrow$. By Lemma \[lem2.4quatro\] we get a contradiction. Thus $\overline{\varphi}_\infty(\ud{x})$ is well-defined for any $\ud{x}\in\tA^{\Z_+}$.

To 