(Purdue Jan ’08) Let \(p:E\to B\) be a covering map. Suppose that points are closed in \(B\). Let \(A\subset E\) be compact. Prove that for every \(b\in B\), the set \(A\cap p^{-1}(b)\) is finite.
Solution
Fix \(b\in B\). Since points are closed in \(B\) and \(p\) is continuous, the fiber \[ p^{-1}(b) \] is closed in \(E\). Hence \[ A\cap p^{-1}(b) \] is closed in the compact space \(A\), so it is compact.
The fiber is also a discrete subspace of \(E\). Indeed, if \(e\in p^{-1}(b)\), choose an evenly covered neighborhood \(U\ni b\), and let \(V\) be the sheet containing \(e\). Since \(p|_V:V\to U\) is a homeomorphism, \[ V\cap p^{-1}(b)=\{e\}. \] Thus each point of the fiber is isolated in the fiber, and the same is true of its subspace \(A\cap p^{-1}(b)\).
A compact discrete space is finite: its cover by singletons has a finite subcover. Therefore \[ A\cap p^{-1}(b) \] is finite for every \(b\in B\).