cluster point of $\{\overline{\varphi}_n(\ud{x})\}_n$ is smaller then any cluster point of $\{\overline{\varphi}_n(\ud{x}^\prime)\}_n$. In particular, if $\overline{\varphi}_\infty$ is well-defined on $\tA^{\Z_+}$, then $\overline{\varphi}_\infty$ is order-preserving.

Let $\ud{x}\prec\ud{x}^\prime$ with $x_k=x^\prime_k$, $k=0,\ldots,m-1$ and $x_m\not=x^\prime_m$. We have $$\overline{\varphi}_{m+n}(\ud{x})=
\overline{\varphi}_m(x_0,\ldots,x_{m-1}+
\overline{\varphi}_n(\sigma^m\ud{x}))\,.$$ By , if $\delta(x_0\cdots x_{m-1})=1$, then $x_m<x^\prime_m$ and for any $n\geq 1$, $\ell\geq 1$, $$\overline{\varphi}_n(\sigma^m\ud{x})=\overline{\varphi}_1(x_m+
\overline{\varphi}_{n-1}(\sigma^{m+1}\ud{x}))\leq
\overline{\varphi}_\ell(\sigma^m\ud{x}^\prime)=
\overline{\varphi}_1(x^\prime_m+
\overline{\varphi}_{\ell-1}(\sigma^{m+1}\ud{x}^\prime))\,;$$ if $\delta(x_0\cdots x_{m-1})=-1$, then $x_m>x^\prime_m$ and $$\overline{\varphi}_n(\sigma^m\ud{x})=\overline{\varphi}_1(x_m+
\overline{\varphi}_{n-1}(\sigma^{m+1}\ud{x}))\geq
\overline{\varphi}_\ell(\sigma^m\ud{x}^\prime)=
\overline{\varphi}_1(x^\prime_m+
\overline{\varphi}_{\ell-1}(\sigma^{m+1}\ud{x}^\prime))\,.$$ Therefore, in both cases, for any $n\geq 1$, $\ell\geq 1$, $$\overline{\varphi}_{m+n}(\ud{x})\leq
\overline{\varphi}_{m+\ell}(\ud{x}^\prime)\,.$$

\[lem2.4ter\] Let $\ud{x}\in\tA^{\Z_+}$ and $x_0=j$.\
1) Let $\delta(j)=1$ and $y_\uparrow(\ud{x})\in \overline{I}_j$ be a cluster point of $\{\overline{\varphi}_n(\ud{x})\}$. Then $f_j\big(y_\uparrow(\ud{x})\big)\geq y_\uparrow(\sigma\ud{x})$ if $y_\uparrow(\ud{x})=a_j$, $f_j\big(y_\uparrow(\ud{x})\big)\leq
y_\uparrow(\sigma\ud{x})$ if $y_\uparrow(\ud{x})=a_{j+1}$ and $f_j\big(y_\uparrow(\ud{x})\big)=y_\uparrow(\sigma\ud{x})$ otherwise. The same conclusions hold when $y_\downarrow(\ud{x})$ is a cluster point of $\{\overline{\varphi}_n(\ud{x})\}$.\
2) Let $\delta(j)=-1$ and $y_\uparrow(\ud{x})\in \overline{I}_j$ be a cluster point of $\{\overline{\varphi}_n(\ud{x})\}$. Then $f_j\big(y_\uparrow(\ud{x})\big)\leq y_\downarrow(\sigma\ud{x})$ if $y_\uparrow(\ud{x})=a_j$, $f_j\big(y_\uparrow(\ud{x})\big)\geq
y_\downarrow(\sigma\ud{x})$ if $y_\uparrow(\ud{x})=a_{j+1}$ and $f_j\big(y_\uparrow(\ud{x})\big)=y_\downarrow(\sigma\ud{x})$ otherwise. The same conclusions hold when $y_\downarrow(\ud{x})$ is a cluster point of $\{\overline{\varphi}_n(\ud{x})\}$.

Set $f_j(\overline{I}_j):=[\alpha_j,\beta_j]$. Suppose for example that $\delta(j)=-1$ and that $n_k$ is the subsequence of all $m$ such that $\delta(x_0,\ldots,x_m)=1$. Since $\delta(j)=-1$ the sequence $\{\oph_{n_k-1}(\sigma\ud{x})\}_k$ is decreasing. Hence by continuity $$\label{mon}
y_\uparrow(\ud{x})=\lim_k\overline{\varphi}_{n_k}(\ud{x})=
\overline{\varphi}(j+\lim_k\overline{\varphi}_{n_k-1}(\sigma\ud{x}))=
\overline{\varphi}(j+y_\downarrow(\sigma\ud{x}))\,.$$ If $y_\uparrow(\ud{x})=a_j$, then $f_j(a_j)=\beta_j\leq
y_\downarrow(\sigma\ud{x})$; if $y_\uparrow(\ud{x})=a_{j+1}$, then $f_j(a_{j+1})=\alpha_j\geq y_\downarrow(\sigma\ud{x})$; if $a_j<y_\uparrow(\ud{x})<a_{j+1}$, 