hyperbolic geometry proof with parallel lines We are assuming hyperbolic geometry in this proof.
Prove that for every line $l$ and external point P (im assuming point $P$ is not on line $l$), there are an infinite number of distinct lines through $P$ parallel to $l$.
I'm thinking about the critical number $r_{0}$ of $P$ and $l$.
My attempt at the proof:
Let $k$ be the intersecting set for $P$ and let $A,B$ be points on $l$ such that line $AB$ is on $l$. If $r\in k$, then $s\in k$ for every $0<s<r$ and there exist $t\in k$ such that $t\in r$. $r_{0}$ is the critical number for $P$ and line $AB$ such that $k$ is the half open interval $[0,r_{0})$ By definition, there exist a point $D$ external to line $AB$, where $D$ is the angle of parallelism (i.e. $\angle APD$) Because we are in hyperbolic geometry, any line parallel to $P$ is also parallel to $l$. If a line is perpendicular to $P$, it will be parallel to $l$ by definition of hyperbolic geometry. Therefore, there are an infinite number of distinct lines through $P$ on line $l$. $QED$
But I feel like this proof is not exactly what is needed to prove that there are an infinite number of lines...
 A: I found your proof not very clear, (k is a set of lines trough P , but what are s and r?)
Was thinking about a simpler proof but came accross quite a lot of unexpected hyperbolic geometry facts.
My first idea was to construct a line that crosses l and both asymtopic parallels but that line does not always exist so I had to come up with another idea. 
Let start.
Theorems used:
(need to be proved seperately)
1) on every segment there are an infinite amount of points
2) two lines cut eachother at at most one point
3) between any two point we can construct a line
Hyperbolic geometry theorems needed
4) trough a point p not on line l there can be drawn two different lines that don't cut line l.
proof 
We have a line $l$ and a point $P$ not on line $l$.
Trough point $P$ we can draw two different lines not intersecting $l$
call these lines ($s$ and $r$ ) 
Take a (real) point $L$ on line $l$ . 
Take a (real) point $R$ on line $r$ not being point $P$
By 3) there is a line trough  $L$ and $R$ call this line $t$.
If line $t$ cuts line $s$ , Point $T$ is at the intersection of $t$ and $s$.
If line $t$ doesn't cut line $s$ , Point $T$ is any point on line t not on ray $RL$  ($R$ is between $T$ and $L$)
For every point on segment $RT$ we can draw a different line trough point $P$
 and none of these lines can cut line l  because they cut lines $s$ and $r$ in point $P$  and to cut line l they need to cut line $s$ or $r$ a second time what is impossible by 2)
Because there are an infinite number of points on segment $RT$ (by 1) there are an infinite number of lines trough P not intersecting line l.
QED
