For a compact covering space, the fibres of the covering map are finite. I am stuck on the following exercise:


  Let $Y$ be a compact topological space, and $p:\ Y\ \longrightarrow\ X$ a covering map. Show that for every $x\in X$ the fibre $p^{-1}(x)$ is finite.


Any help would be very welcome.
The space $X$ has a finite open cover $(U_i)_i$ of evenly covered neighborhoods. We can assume that the cover is minimal, that means none of these sets can be removed. The preimage of each $U_i$ is a disjoint union of open sets $\left(V_i^j\right)_{j\in J_i}$ each of which is mapped homeomorphically onto $U_i$. Since the $U_i$ cover $X$, the $V_i^j$ cover $Y$, so there must be a finite subcover.

By minimality of the cover of $X$, every $U_i$ has an element $x_i$ not contained in any other $U_j$. The finitely many $V_k^l$ cover $Y$, thus cover $p^{-1}(x_i)$. But this fiber is disjoint from any $V_k^l$ with $k\ne i$, so $p^{-1}(x_i)$ must be covered by sets $V_i^1,...,V_i^m$. Since each of them contains only one point of the fiber, it must be finite.

Now since the fibers of all points in $U_i$ have the same cardinality, we conclude that all fibers are finite.
Maybe these results will be helpful:
I'm assuming: Top space is Hausdorff, and singletons are closed in bottom space.

1)The fibers of a singleton are discrete.

2) If a singleton y in your space is closed; $U_y$ is a covering neighborhood of $y$,
with restricition homeomorphism $P_y: V_y \rightarrow U_y$ , then $P^{-y}$ is a closed
set. A closed subset of a compact Hausdorff space is compact. 

3)The only compact discrete spaces are finite spaces.

Can you take it from here?