what would be a euclidean argument for why the characteristic axiom always holds in hyperbolic geometry?

Question

what would be a euclidean argument for why the characteristic axiom always holds in hyperbolic geometry?

Characteristic axiom states Given a line k and a point p not on k, there are at least two lines on p that do not intersect k.

Answer 1

First of all, the axiom is, after all, just presumed to hold -- that is after all the point of axioms.

However, you can get some intuitive grasp on why geometry on a hyperboloid would have this property. Consider a line $\ell$ on the hyperboloid and a nearby exterior point $P$. Now try to make a parallel line through $P$ in a naive way, by taking two points $(A,B)$ very near to $P$ and on opposite sides to the shortest line from $P$ to $\ell$, and choosing then so that the minimum distances of $A$ and $B$ to $\ell$ are equal. Line $AB$ comes very close to $P$; tweak it by increasing the equal distances from $A (B)$ to $\ell$ by equal amounts until the new line passes through $P$.

Now this line $APB$ is actually a geodesic on the hyperboloid, and you will find that as you go further away from $P$, the shortest distance of a point $C$ on $APB$ to the line $\ell$ becomes greater and greater.

Finally, imagine "tilting" this line by an angle at $P$ which is small enough that the rate of approach to $\ell$ due to the tilt (on one side) is less than the rate of increase in the distance due to the concave shape of the hyperboloid. This tilted line will also never meet $\ell$. So you can make at least two lines through $p$ that never meet $\ell$.

It is in fact possible to work out explict equations for these lines, but that doesn't add much to the intuition.

You can prove, without introducing new axioms, that if there are at least $2$ such lines through any point, then there are infinite such lines.