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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 Möbius 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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    Alternatively, the proof in the case of sequential compactness is quite simple.2013-09-13

2 Answers 2

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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!

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    So, that's the way of making the intuition in my flawed argument in the comments into a proof. Alexander's subbase lemma (indeed in Kelley, Thm. 6 of Chapter 5 of the 1955 edition) + the precise use of local triviality was what I was missing. Nice!2012-03-28
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I don't get where the problem is. Am I missing something?

Each point $b\in B$ has a neighbourhood $N_b$ such that the bundle over $N_b$ is trivial. Choose a smaller closed (thus compact) neighbourhood $C_b$ (we need some weak assumption here like Hausdorffness of $B$). The bundle over $C_b$ is homeomorphic to $C_b\times F$, thus compact.

Note that $\{\mathrm{int}(C_b)\}_{b\in B}$ is an open cover of $B$ and by compactness has a finite subcover indexed by $(b_i)_{i=1}^n$. Consequently $E$ is a finite sum of compact sets, thus compact: $E = \bigcup_{1\leq i \leq n} \pi^{-1}(C_{b_i}).$

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    This works. The argument I gave in my answer was intended only to show how to fix t.b.'s idea from the comments to the question. In fact, I'd say that the two arguments are, up to ordering, the same.2013-09-14