Why can we assume the other endpoint of this hyperbolic line lies after $x=1$? (Anderson's Hyberbolic Geometry Prop. 3.22)

Question

I'm reading through James Anderson's Hyberbolic Geometry. Proposition 3.22 and the first step of it's proof is as follows

Proposition 3.22: Let $l_0$ and $l_1$ be ultraparallel hyperbolic lines in $\mathbb{H}$. Then $d_{\mathbb{H}}(l_0, l_1)>0$.

Proof: Again, by making use of the triple transitivity of $\mathrm{Mob}(\mathbb{H})$ on $\overline{\mathbb{R}}$, we may assume that the endpoints at infinity of $l_0$ are 0 and $\infty$, and that the endpoints at infinity of $l_1$ are $1$ and $x > 1$. [Continues]....

I know that triple transitivity of $\mathrm{Mob}(\mathbb{H})$ on $ \overline{\mathbb{R}}$ allows us to send three of the four endpoints to $0,1,\infty$ as such. But I thought that doing so, we will not have any control on the image of the fourth endpoint; it could be anywhere other than those three images. How can we assume that for the image $x$ of the fourth endpoint, we have $x>1$?

I suspect that otherwise we could act some element of $\mathrm{Mob}(\mathbb{H})$ to get the desired result; But I can't see such element of the group. And also I suspected that conformality of elements of $\mathrm{Mob}(\mathbb{H})$ automatically forces $x$ to be greater than $1$, but I can't tell.

Edits: As it seem's that $\mathrm{Mob}(\mathbb{H})$ is not something generally agreed upon, here it means the group of transformations of either of the two following forms:

1. $ \frac{az+b}{cz+d} $ where $a,b,c,d \in \mathbb{R}$ and $ad-bc=1$.
2. $ \frac{a\bar{z}+b}{c\bar{z}+d} $ where $a,b,c,d \in i \mathbb{R}$ and $ad-bc = 1$.

Answer 1

Write the four points in cyclic order as $a,b,c,d$, where $a,b$ are the endpoints of $l_0$ and $c,d$ are the endpoints of $l_1$. Here by "cyclic order" I mean that one of four inequalities holds:

• $a<b<c<d$ (allowing $d=\infty$);
• $b<c<d<a$ (allowing $a=\infty$);
• $c<d<a<b$ (allowing $b=\infty$);
• $d<a<b<c$ (allowing $c=\infty$).

Now choose a Möbius transformation in $\text{PSL}_2(\mathbb R)$ that has the effect $a \mapsto a'=\infty$, $b \mapsto b'=0$, $c \mapsto c'=1$. Because elements of $\text{PSL}_2(\mathbb R)$ respect cyclic order, this forces $d \mapsto d' \in (1,+\infty)$.