# Definition of hyperbolic lenght.

Theorem 1: Let $\text{arc(AB)}$ be an arc of an equidistant curve (Which can be a circle, a horocircle or an equidistant line) and $(A^{n})$ a sequence of partitions of the arc $\text{arc(AB)}$ such that $\lVert A^{n}\rVert \rightarrow 0$ and $l(A^{n}) \rightarrow l$, where $l>0$. Then the lenght of the arc $\text{arc(AB)}$ is $l$.

Obs.: Norm of the partition is defined as the biggest lenght of the segments of the partition and the lenght of a partition is the sum of all the lenghts of the segments of the partition, as usual.

Proof: $l(A^{n}) \leq c, \forall n$, where $c$ is the lenght of the arc $\text{arc(AB)}$, since $c$ is the supremum of the set of all lenghts of partitions. Taking the limit we have $l \leq c$.

Let $(B)$ be an arbitrary partiton of $\text{arc(AB)}$. Let $\{C^{n}\} = \{A_{i}^{n}\} \cup \{B_{j}\}$. Then, as $C^{n}$ refines $(B)$, $l(B) \leq l(C^{n})$. If $n$ is big enough, as $\lVert A^{n}\rVert \rightarrow 0$ , we can assure that each arc $\text{arc}(A_{i-1}A_{i})$ has at most one point of $(B)$.

(The proof continues, but I want to make a stop here).
I was understanding everything until this last information. How can I be sure that there will be at most one point?

The autor says that it’s related to this exercise:

Let $C$ an equidistant curve and $A,B \in C$. Show that there is a constant $k = k(C,A,B)>0$ such that for all $D,E \in C$ it is valid: If $A,B \in \text{arc(D,E)}$, then $DE \geq k$

But I can’t see why there is a relation.