On a pair of pants, complete geodesics do not cover the whole surface I am reading a thesis by Jenya Sapir. In her thesis, she mentions that "on a pair of pants, complete geodesics no longer cover the whole surface". 
I could not figure out how can we come up with this claim but I knew that, when the length of each geodesic at the boundary of the surface is long enough, then this claim is true obviously. 
Further questions, whether this property holds when we consider any hyperbolic surface with boundary?
Thank you in advance!
 A: I do not think it is clear if the claim is true, I suggest to write to Sapir and ask for details. (It is also not entirely clear what it is claimed, namely if she means "for some" hyperbolic pairs --- in which case the claim is proven below of pants or "for all" --- in which case the claim is unclear.) 
Here is what one can say in general. Let $S$ be a connected metrically complete hyperbolic surface with nonempty geodesic boundary and of finite area. Then $S=C/\Gamma$ where $C$ is a convex subset in ${\mathbb H}^2$ (with nonempty geodesic boundary) and $\Gamma< Isom({\mathbb H}^2)$ is a discrete subgroup preserving $C$. Let $\Lambda$ denote the limit set of $\Gamma$: $C$ is the closed convex hull of $\Lambda$. Let $\Delta$ denote the diagonal in $\Lambda \times \Lambda$. There is a map
$$
f: (\Lambda \times \Lambda - \Delta)\times {\mathbb R}\to C, 
$$
sending a pair of distinct points $\lambda_1, \lambda_2$ in $\Lambda$ to a (parameterized with the unit speed) oriented geodesic in the hyperbolic plane asymptotic to $\lambda_1, \lambda_2$. (Fix a base-point $o\in {\mathbb H}^2$ and send $(\lambda_1,\lambda_2,0)$ to the point on the hyperbolic geodesic spanned by $\lambda_1,\lambda_2$ which is closest to $o$. Then extend to the rest of $(\lambda_1,\lambda_2,t)$ in the obvious fashion.) 
The question is if this map is non-surjective. (It is equivalent to asking if $S$ is not covered by biinfinite geodesics.) 
One can work, for instance, with the Klein model of the hyperbolic plane and check that the map $f$ is (locally) Lipschitz (the target is equipped with the restriction of the Fubini-Study metric on $RP^2$). Then the Hausdorff dimension of the domain  of the map is $2d+1$ where $d$ is the Hausdorff dimension of $\Lambda$. The target, of course, has Hausdorff dimension $2$, so the map is not surjective if $d<1/2$ since Lipschitz maps cannot increase Hausdorff dimension. However, if $S$ has finite area, it can be always equipped with a different metric, for which $d$ is as close to $1$ as you like (but, necessarily, $<1$). Similarly, one can always find a metric with $d$ as close to $0$ as you like. 
The domain of the map $f$ has topological dimension $1$. There is a common fallacy that Lipschitz maps cannot increase topological dimension. (In fact, they can.) A worry is that the cited claim relies upon this common fallacy.  
Edit.  Here is a proof that $dim(\Lambda\times \Lambda \times {\mathbb R})= 2d+1$. In general, if $X, Y$ are (separable) metric spaces (in our case, these are subsets of Euclidean spaces) then 
$$
dim(X)+dim(Y) \le dim(X\times Y)\le dim(X) + dim_B(Y),
$$
where $dim_B$ is the box-counting dimension. See for instance, here. In particular, if $dim_B(Y)=dim(Y)$ then 
$$
dim(X\times Y)=dim(X) + dim(Y). 
$$
In our case, we are multiplying $\Lambda$ (twice) and ${\mathbb R}$. For ${\mathbb R}$, of course, $dim=dim_B$. If $\Gamma$ is a geometrically finite Kleinian group then $dim(\Lambda)=dim_B(\Lambda)$, see for instance
B.Stratmann, M.Urbanski, The box-counting dimension for
geometrically finite Kleinian groups, Fund. Math, 1996. 
(More general results were proven by Bishop and Jones.) I am not sure what happens for squares of limit sets of geometrically infinite groups but you probably do not care. 
