Universal Cover of a Surface with Boundary. What does Cantor set on Boundary Correspond to? I am trying to understand in more detail the answer to:
Universal Cover of a Surface (with Boundary)
It is mentioned that the universal cover of a hyperbolic surface $S$ with  geodesic boundary is a closed disk $D^2$ with a Cantor set removed from its boundary. I am trying to see what the preimage of an interval of the Cantor set is under this cover, i.e., why are we removing the Cantor set from the boundary . I understand we first construct the surface $S$ by gluing pairs of pants and then  embed $S$ in a   closed hyperbolic surface $S^$ with boundary, and that the universal cover $S^ ~$ (which we construct by gluing more pants so that we cap all the boundar components.) is the Poincare disk
$D^2:= \mathbb H^2 \cup S^1_{\infty} $ (i.e., the last part is the x-axis plus the point at $\infty$ .) But where does the removed Cantor set in the cover  of the surface with boundary come from? Are we getting a sort-of infinitely-sheeted cover, and, if so, is there a reasonably-nice expression for the projection map? And if someone had a ref. for explaining why the lift of a hyperbolic surface without boundary is the Poincare disk, that would be great.
 A: The short answer to your question is that the Cantor set corresponds to the limit set of the action of $\pi_1 S$ on $\mathbb{H}^2$.
Here are more details, the idea being that the topological description of the universal cover $\tilde S$ is a consequence of a more detailed geometric description which can be found in any basic textbook on hyperbolic manifolds, such as the book of Thurston et.al. or the book of Ratcliffe.
The first basic fact one needs is that the universal cover $\tilde S$ of $S$ with its lifted hyperbolic structure embeds isometrically in $\mathbb{H}^2$, so that the deck transformation action of $\pi_1(S)$ on $\tilde S$ extends to a free and properly discontinuous action of $\pi_1(S)$ on all of $\mathbb{H}^2$ by isometries.
Next, one needs the concept of the limit set $\Lambda$ of the action of $\pi_1(S)$ on $\mathbb{H}^2$, which in this setting is equal to the set of points in the circle at infinity $S^1_\infty = \partial \mathbb{H}^2$ which are contained in the closure of $\widetilde S$ relative to the closed disc $\overline{\mathbb{H}}^2 = \mathbb{H}^2 \cup S^1_\infty$. This closure of $\tilde S$ is equal to the union $\tilde S \cup \Lambda$, and is also equal to the convex hull of the set $\Lambda$ which means the smallest closed subset of $\overline{\mathbb{H}}^2$ which contains $\Lambda$ and contains the hyperbolic geodesic segment between any two of its points. It follows from this description that $\tilde S \cup \Lambda$ is homeomorphic to the closed 2-disc, and it also follows that $\Lambda$ is on the boundary of that 2-disc.
What remains, then, is to see that $\Lambda$ is homeomorphic to the Cantor set. One shows directly, using the fact that $\Lambda$ is the limit set of the action $\pi_1(S)$, that $\Lambda$ is perfect and totally disconnected. Then one applies the basic theorem that every perfect, totally disconnected, nonempty, compact metric space is homeomorphic to the Cantor set.

Let me also correct a few mis-impressions in your question. 
As said in my answer above, the universal cover $\tilde S$ is embedded in $\mathbb{H}^2$ which is the interior of the Poincare disc. But $\tilde S$ is a proper subset of the interior of the Poincare disc, so $\tilde S$ is neither equal to the whole of the Poincare disc (because $\tilde S$ is disjoint from the boundary of the Poincare disc), nor is it equal to the whole of the interior of the Poincare disc.
Also, In this process you should not cap off the boundary components of $S$. Each component of $\partial S$ is a geodesic circle, and each component of $\partial \tilde S$ is a geodesic line in $\mathbb{H}^2$. There is a countably infinite number of components of $\partial \tilde S$. The universal covering map $\tilde S \to S$ restricts to a universal covering map from each component of $\partial \tilde S$ to some component of $\partial S$. 
A: This is a mild reinterpretation of the last part of the answer to the question you link and not a detailed answer:
The hyperbolic geometry argument given in that answer is equivalent to completing $S$ to a hyperbolic surface with cusps, taking the universal cover of that (which is the entire hyperbolic plane, by uniformization), and then realizing $S$ as the quotient of a subset of that universal cover (i.e. by deleting neighborhoods of the cusps).
The Cantor set mentioned is the "gaps" along the boundary between the preimages of the cusps. See Figure 1 in Honda-Kazez-Matic which we were discussing in a previous question for a picture of this cover. The Cantor set in that picture corresponds to the open parts of the boundary there (i.e. the parts of the boundary that preimages of the neighborhood of the cusp didn't meet).
