such that $z_i \in [p, \nu]$ from some step.

If $z_i \to p$ with $p \in {\mathscr{Z}}$, then $z_i = p$ from some step by Remark \[R:dichotomy\], so $z_i \to p$ from below [*and*]{} from above.

\[Infimum and supremum\] Let $a,b\in S^1$. Each non-empty subset $P \subseteq [a,b] \cap {\mathscr{Z}}\subset S^1$ has an infimum and a supremum, and there is a decreasing sequence in $P$ converging to its infimum, denoted by $\inf_{[a,b]} P$, and an increasing sequence in $P$ converging to its supremum, denoted by $\sup_{[a,b]} P$.

Note that the infimum and the supremum are contained in the interval $[a,b]$, but not necessarily in $P$ or in ${\mathscr{Z}}$. If $i = \inf_{[a,b]} P$ and $s=\sup_{[a,b]} P$ then $a \leqslant i \leqslant p \leqslant s \leqslant b$ for each $p \in P$.

Note that any increasing or decreasing sequence in an interval $[a,b]$ is convergent to a point in that interval.

Recall that ${\mathscr{Z}}$ contains infinitely many points by Definition \[D:admissible\](i). For each $z \in {\mathscr{Z}}$, the sequence $\{ z^{+n} \}_{ n \geqslant 0 }$ defined iteratively by $z^{ +0 } = z$ and $z^{ +(k+1) } = ( z^{ +k } )^+$ for each $k \in {\mathbb{Z}}_{\geqslant 0}$ is an 