increasing sequence. Moreover, there are infinitely many points of ${\mathscr{Z}}$ in $[z,z^-]$ whence $z \leqslant z^{ +n } < z^-$. So $\{z^{+n}\}_{n \geqslant 0}$ is an increasing sequence in $[z,z^-]$ and it must converge to a limit point.

The limit point of $\{z^{+n}\}_{n \geqslant 0}$ will be denoted $z^{ +\infty }$. Symmetrically, we can define $\{ z^{ -n } \}_{n \geqslant 0}$ and its limit point will be denoted $z^{ -\infty }$.

\[L:limit\] We have $[z,z^{+\infty}] \cap L({\mathscr{Z}}) = \{z^{+\infty}\}$ and $[z^{-\infty},z] \cap L({\mathscr{Z}}) = \{z^{-\infty}\}$.

We only prove that $[z,z^{+\infty}] \cap L({\mathscr{Z}}) = \{z^{+\infty}\}$; the equality $[z^{-\infty},z] \cap L({\mathscr{Z}}) = \{z^{-\infty}\}$ is proved symmetrically. The inclusion $\{z^{+\infty}\} \subseteq [z,z^{+\infty}] \cap L({\mathscr{Z}})$ is clear by definition. The inclusion $[z,z^{+\infty}] \cap L({\mathscr{Z}}) \subseteq \{z^{+\infty}\}$ amounts to showing that $[z,z^{+\infty}) \cap L({\mathscr{Z}}) = \varnothing$, which again amounts to showing that $(z,z^{+\infty}) \cap L({\mathscr{Z}}) = \varnothing$, since $z\in {\mathscr{Z}}$, and hence $z \notin L({\mathscr{Z}})$. So suppose for a contradiction that there exists $x\in (z,z^{+\infty}) \cap L({\mathscr{Z}})$. In particular, $x\notin{\mathscr{Z}}$ and there exists a sequence $\{z_i\}_{i \in {\mathbb{Z}}_{\geqslant 0}}$ from ${\mathscr{Z}}$ converging to $x$. By construction we have that $z^{+\infty} = \sup_{[z,z^-]}\{z^{+n} \mid n \geqslant 0\}$, so we can find $m \geqslant 0$ such 