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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.

azimut
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Alex Becker
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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 04:28
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    @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-Álvarez Mar 28 '12 at 04:34
  • @t.b. I tried something quite similar to that, yet ran into the problem Mariano pointed out. – Alex Becker Mar 28 '12 at 04:36
  • @Mariano: You're right. That was quite silly, thanks for the correction. – t.b. Mar 28 '12 at 04:36
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    @t.b., the natural thing to do is never silly. :) – Mariano Suárez-Álvarez Mar 28 '12 at 16:42
  • Alternatively, the proof in the case of sequential compactness is quite simple. – Ryan Budney Sep 13 '13 at 18:20

2 Answers2

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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!

Mariano Suárez-Álvarez
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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! – t.b. Mar 28 '12 at 04:58
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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: $$E = \bigcup_{1\leq i \leq n} \pi^{-1}(C_{b_i}).$$

savick01
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  • Why the downvote? If I am missing something, I would like to know. – savick01 Sep 13 '13 at 22:05
  • I did not downvote, but I don't think that Hausdorff is sufficient for the property you use. Consider the set $[0,\omega]$ with the order topology, which is compact and Hausdorff but every point except $\omega$ is open. – Alex Becker Sep 14 '13 at 00:59
  • @AlexBecker It is, but I don't get why it is a problem. My argument is as follows: consider the complement $N_b^c$ of $N_b$ - it is closed, so compact. Thus (by Hausdorffness) there are disjoint neighbourhoods of $b$ and $N_b^c$. I take $C_b$ equal to the complement of that neighbourhood of $N_b^c$. – savick01 Sep 14 '13 at 01:19
  • Ah, I see the argument. What was tripping me up is that $C_b$ need not be strictly smaller than $N_b$; in the example I gave for all $b\ne \omega$ one could have $C_b=N_b$. – Alex Becker Sep 14 '13 at 02:37
  • 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. – Mariano Suárez-Álvarez Sep 14 '13 at 12:13