# Given a fiber bundle $F\to E\overset{\pi}{\to} B$ such that $F,B$ are compact, is $E$ necessarily compact?

Consider a (locally trivial) fiber bundle $F\to E\overset{\pi}{\to} B$, where $F$ is the fiber, $E$ the total space and $B$ the base space. If $F$ and $B$ are compact, must $E$ be compact?

This certainly holds if the bundle is trivial (i.e. $E\cong B\times F$), as a consequence of Tychonoff's theorem. It also holds in all the cases I can think of, such as where $E$ is the Mobius strip, Klein bottle, a covering space and in the more complicated case of $O(n)\to O(n+1)\to \mathbb S^n$ which prompted me to consider this question. I am fairly certain it holds in the somewhat more general case where $F,B$ are closed manifolds. However, I can't seem to find a proof of the general statement. My chief difficulty lies in gluing together the local homeomorphisms to transfer finite covers of $B\times F$ to $E$. Any insight would be appreciated.

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How about this: Take an open cover of $E$. For each point $b \in B$ choose a finite subcover $U_{b,1}, \ldots, U_{b,n(b)}$ of the fiber $\pi^{-1}(b)$ over $B$. Project those sets down and intersect them to get an open set $V_{b}$. Then $\pi^{-1}(V_b)$ is contained in their union. The sets $V_b$ cover $B$, so we find $b_1, \ldots, b_n$ such that $V_{b_1},\ldots,V_{b_n}$ covers $B$. The collection $U_{b_j,1},\ldots, U_{b_j,n(b_j)}$, $j=1,\ldots,n$ is a finite subcover of the cover we started with. – t.b. Mar 28 '12 at 4:28
@t.b.it may well be the case that your $V_b=B$ yet $E$ is not contained in the union of $U_{b,1},\dots,U_{b,n(b)}$. – Mariano Suárez-Alvarez Mar 28 '12 at 4:34
@t.b. I tried something quite similar to that, yet ran into the problem Mariano pointed out. – Alex Becker Mar 28 '12 at 4:36
@Mariano: You're right. That was quite silly, thanks for the correction. – t.b. Mar 28 '12 at 4:36
@t.b., the natural thing to do is never silly. :) – Mariano Suárez-Alvarez Mar 28 '12 at 16:42

By local triviality, there is a open covering $\mathcal U$ of $B$ such that for each $U\in\mathcal U$ the open subset $\pi^{-1}(U)$ of $E$ is homeomorphic to $U\times F$ in a way compatible with the projection to $U$. It follows that there is a subbase $\mathcal S$ of the topology of $E$ consisting of open sets each of which is contained in one of these $\pi^{-1}(U)$ and corresponding under those homeomorhisms to an open subset of $U\times F$ of the form $V\times W$ with $V\subseteq U$ open in $B$ and $W\subseteq F$ open in $F$.
To prove compactness, it is enough to show that every covering of $E$ by subsets of $\mathcal S$ contains a finite subcovering —this is called Alexander's subbase lemma and is used in one of the proofs of Thychonof's theorem (for example, in Kelley's book, iirc). Do that!