# If $p:E\to B$ is a covering space and $p^{-1}(x)$ is finite for all $x \in B$, show that $E$ is compact and Hausdorff iff $B$ is compact and Hausdorff

I can show that if $E$ is compact and Hausdorff $B$ has the same properties, also I can show that if $B$ is compact and Hausdorff $E$ is Hausdorff, but I have troubles trying to prove that $E$ is also compact. Any suggestions would be appreciated.

I would like to know if there is a short way or at least a simple way to show that if E is Hausdorff so is B, I can prove it but I have to make a lot of observations and I get a really really long demostration.

This is an exercise in Hatcher (Algebraic Topology) Section 1.3, exercise 3

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You might also want to try doing the exercise making extensive use of ultra filters. Although they are not needed here, the resulting proof can be made much shorter. –  Alexander Thumm Jun 12 '12 at 5:10

Let $\mathcal{U}$ be an open cover of $E$. Then for each $x\in B$ there exist $p^{-1}(x)$ is finite. Thus we can choose $U^x_1,\ldots, U^x_{n_x}\in\mathcal{U}$ such that $p^{-1}(x)$ is in the union of these sets.
Hints: Look at the image of $U^x_1,\ldots,U^x_{n_x}$ under $p$. Can you get an open set of $B$ from this containing $x$? How can you use this to get an open cover of $B$? How do you extract an open cover of $E$ from this information?
Well, For any $x \in B$ I can get an evenly covered neighborhood $U_{x}$ so its inverse image is a finite union of open sets homeomorphic to $U_{x}$, $\bigcup_{k=1}^{n_{x}} A_{k}$; for each $k$, there is an unique element $y_{k} \in A_{k}$ such that $p(y_{k})=x$. For each $y_{k}$ there is also an open set $S_{k}$ in the cover with $y_{k} \in S_{k}$, so restricting $p$ to $S_{k} \cap A_{k}$ i get an homeomorphism between $S_{k} \cap A_{k}$ and an open set contained in $U_{x}$. –  Frank Jun 12 '12 at 4:12
By taking direct image of these open sets I can get an open cover of $B$, so I can take a finite subcover, the problem is how to extract t open cover of E when I take inverse image of this finite subcover? Do I have to modify the cover first? –  Frank Jun 12 '12 at 4:12
For a fixed $x\in B$, consider the images $p(S_k\cap A_k)$. As you have said these are all open in $B$. What do you know about their intersection (I repeat, for the moment, $x$ is fixed)? Is it open? Is it nonempty? If we call this intersection $V_x$, what can we say about its preimage, $p^{-1}(V_x)$? –  J. Loreaux Jun 12 '12 at 5:21
The intersection is open and obviously not empty because $x$ belongs to all of them. So with those intersections I create the open cover for $B$ right? Then when I take the inverse image of each open set in the finite subcover I get a finite union of open sets and each one is contained in at least one element of the initial cover. Finally I have obtained my finite subcover! Thanks a lot! –  Frank Jun 12 '12 at 5:30