Arcs and surfaces. Why are there finitely many arcs on the surface up to the action of MCG? Given a bordered surface $S$ (I imagine this is true for non-orientable surfaces too, but you may restrict to the case of orientable surfaces) with finitely many marked points on each boundary component.
Is it true that up to the action of the mapping class group $MCG(S)$ there are only finitely many arcs connecting marked points?
 A: Yes, it is true. In brief, the reason is that there are only finitely many possibilities for the surface $S \setminus \alpha$ that is obtained by cutting $S$ along $\alpha$. Let me give some details.
Consider such an arc $\alpha$. Cutting along $\alpha$, let $S \setminus \alpha$ denote the resulting bordered surface, which inherits marked points on its boundary from the marked points on the boundary of $S$, and which has two special arcs on its boundary obtained from cutting along $\alpha$. Let's put "gluing decorations" on $S \setminus \alpha$, by coloring those two arcs red and matching their orientations in pairs according to the orientations of $\alpha$. Thus, the surface $S$ and the arc $\alpha$ can be recovered from $S \setminus \alpha$ as a quotient using the gluing decorations.
Let me now assume an important missing hypothesis, namely that the surface $S$ is compact and connected. It follows $S \setminus \alpha$ is also compact, it has at most two components, and one obtains bounds on the Euler characteristics and on the numbers of marked points of its components, bounds which are derived from the Euler characteristic and number of marked points of $S$ itself.
From this, and by applying the classification of surfaces, you can deduce that the surface $S \setminus \alpha$ falls into one of a specific finite set of decoration preserving homeomorphism classes relative to marked points. Let $N$ be the number of such classes.
Now we can conclude that, up to the action of $MCG(S)$, there are only $N$ arcs connecting marked points, because for any $N+1$ such arcs, say $\alpha_1,\ldots,\alpha_{N+1}$, amongst the surfaces $S - \alpha_1,\ldots,S-\alpha_{N+1}$ the pigeonhole principle will produce two of them, say $S - \alpha_i$ and $S-\alpha_j$, for which there exists a decoration preserving homeomorphism $f : S - \alpha_i \to S - \alpha_j$, and so $f$ induces a homeomoprhism $S \to S$ taking $\alpha_i$ to $\alpha_j$. 
