About covering maps! Can someone post a proof of the statement that if $X$ is compact then the covering map $q:E\rightarrow X$ is finitely sheeted given that $E$ is compact as well. 
 A: Suppose $p$ is an infinite sheeted cover. Let $x\in X$ and let $U_x$ be a neighbourhood of $x$ such that $p^{-1}(U_x)$ is homeomorphic to a disjoint union of infinitely many copies of $U_x$. Index these subsets by some infinite indexing set $I$ so $$p^{-1}(U_x)\cong\bigsqcup_{i\in I}V^{(i)}_x$$ where $V^{(i)}_x$ is homeomosphic to $U_x$ for all $i\in I$. Further, suppose that the restriction of $p$, $p|V_{x}^{(i)}\colon V_{x}^{(i)}\rightarrow U_x$ is a homeomorphism. Such a set $U_x$ is guaranteed by the definition of a covering space.
Note that the collection of sets $\{V^{(i)}_x \mid \forall x\in X,\forall i\in I\}$ is a cover for $E$ and so has a finite subcover. Suppose such a finite subcover is given by the set $A=\{V^{(i_0)}_{x_0},\ldots, V^{(i_n)}_{x_n}\}$. Now, the open set $V^{(i_0)}_{x_0}$ only covers a single point in the fiber of the point $x_0$ and, because $A$ is finite, there exists a $k$ such that $V_{x_k}^{(i_k)}$ covers an infinite number of points in the fiber of $x_0$.
This is clearly a contradiction however, as the definition of a covering map says that $p$ restricted to any one of the homeomorphic copies of $U_x$ in the preimage of $U_x$ is itself a homeomorphism. But $p|{V_{x_k}^{(i_k)}}$ isn't a homeomorphism because it is not injective (an infinite number of points in $V_{x_k}^{(i_k)}$ get mapped to $x_0$). We conclude that $p$ is not infinite-sheeted.
