# Fiber bundle is compact if base and fiber are

I want to show that the total space $E$ is compact if the fiber $F$ and the base space $B$ are compact. Let $\pi$ denote the fiber projection. Since every point in $B$ has an open neighborhood $U$ whose preimage $\pi^{-1}(U)$ is homeomorphic to $U\times F$ via $\phi_U=(\pi,\beta_U)$, there are finitely many such sets covering $B$. Let's call this collection $\mathcal U$. Then $\mathcal S:=\{\phi_U^{-1}(V\times W)\mid U\in\mathcal U,V\text{ open}\subseteq U,W\text{ open}\subseteq F\}$ is a subbase for the topology on $E$.

By Alexander's subbase lemma we can restrict ourselves to considering a cover $\mathcal C\subseteq\mathcal S$ of $E$. (Mariano Suárez-Alvarez recommends this in his answer here) For each $b\in B$ the fiber $\pi^{-1}(b)$ is compact and thus contained in the union ${\bigcup_{i=1}^{n_b}\phi^{-1}_{U_i^b}(V_i^b\times W_i^b)}$ of sets in $\mathcal C$. Define $V^b:=\bigcap_{i=1}^{n_b}V_i^b$. By compactness $B=\bigcup_{k=1}^l V^{b_k}$. I think that $E=\bigcup_{k=1}^l\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$, similar to the proof of compactness of the product of two compact spaces.

Unlike the trivial bundle case, however, where the homeo's are just identities, here the $\phi_{U}$'s could be pretty ugly, especially where they are overlapping, and we don't have $y\in\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})\$ whenever $\ \pi(y)\in V^{b_k}$

So I try to get to a contradiction by assuming that for each $k=1,\dots,m$ such that $\pi(y)\in V^{b_k}$ we have $\forall i=1,\dots,n_k:\ y\notin\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$ which then implies $\beta_{U^k_i}(y)\notin W_i^{b_k}$. If there were a $k$ such that $U^k_1,\dots,U^k_{n_k}$ were all the same then I would be finished. But maybe the $b_k$'s should have been chosen in certain way.

Maybe someone remembers how the proof worked.

• It would be so much easier to do this for sequential compactness. – A Blumenthal Mar 7 '13 at 16:12
• @ABlumenthal. It would. But the spaces are general topological spaces. – Stefan Hamcke Mar 7 '13 at 16:16
• @ABlumenthal. But one could use nets or filters. Is that an alternative? – Stefan Hamcke Mar 7 '13 at 16:18
• I was just thinking that- a space is compact iff every infinite net has a convergent subnet. – A Blumenthal Mar 7 '13 at 16:20
• @ABlumenthal: Or, equivalently, iff every net has a cluster point. I'm going to try that. I just decided to try the subbase cover due to Mariano's answer here. – Stefan Hamcke Mar 7 '13 at 16:24

We use the following characterization of compactness: every infinite net has a convergent subnet.

Let $\{x_{\lambda}\}_{\Lambda}$ be an infinite net in $E$, $\Lambda$ the directed set of the net. As $B$ is compact, there is an infinite subnet $\Lambda_1 \subset \Lambda$ for which $\{\pi(x_{\lambda})\}_{\Lambda_1}$ converges to a point $b \in B$. Fix a local trivialization $\phi_U : \pi^{-1}(U) \rightarrow U \times F$ for $b \in U$. Pare off the net $\Lambda_1$ so that $\pi(x_{\lambda}) \in U$ for all $\lambda$.

Now let $\pi_2 : U \times F \rightarrow F$ be the continuous projection onto the second factor. So, $\{\pi_2 \circ \phi_U(x_{\lambda})\}$ is an infinite net in $F$, hence possesses yet another convergent subnet $\Lambda_2$. Then $\{\phi_U(x_{\lambda})\}_{\Lambda_2}$ is convergent in $U \times F$ by definition of the product topology, and the image of this convergent net under the homeomorphism $\phi_U^{-1}$ is convergent. Thus $\{x_{\lambda}\}_{\Lambda_2}$ is a convergent subnet.

• That so much easier than the "open cover"-method. I guess a proof via (ultra-)filters could not be any simpler either? Thanks a lot! – Stefan Hamcke Mar 7 '13 at 16:58
• @StefanH. I think the moral of the story is that the fiber bundle structure is very nice locally, and local methods prevail in that situation, whereas these open cover methods are too global for the fiber bundle to say anything without some effort. In a similar vein, it might be a nice exercise to prove the compactness of $E$ using the finite intersection property... – A Blumenthal Mar 8 '13 at 19:15

I came across (while I was actually searching for something else) another nice proof which makes use of several facts which are also useful on their own. I'm going to post it here:

Lemma 1: If $F$ is compact, then the projection $p:F\times Y\to Y$ is closed.
Proof: Let $C$ be closed and $y\notin p(C)$. Then $\{y\}\times F$ is covered my finitely many open product sets, each of which does not intersect $C$. The intersection of their projections is an open neighborhood of $y$ not intersecting $p(C)$. Thus $p(C)$ is closed.

Lemma 2: Let $\{A_i\}$ be a family of subsets of a topological space $Y$ such that $\{\mathrm{int}A_i\}$ covers $Y$. A subset $B$ is closed if $B\cap A_i$ is closed in $A_i$ for every $i$.
Proof: Let $B\subseteq Y$ such that $B\cap A_i$ is closed in $A_i$ for every $i$. Let $D=Y-B$. Since $D\cap A_i$ is open in $A_i$, thus $D\cap\mathrm{int}A_i$ is open in $Y$. Their union equals $D$, hence $B$ is closed.

Lemma 3: Let $f:X\to Y$ be a function, where $Y$ is covered as in lemma 2. Then $f$ is closed if the restrictions $f_i:f^{-1}(A_i)\to A_i$ are closed for all $i$.
Proof: Let $C$ be closed in $X$. Then $f(C)\cap A_i=f_i(C\cap f^{-1}(A_i))$ which is closed in $A_i$ by hypothesis. By lemma 2 $f(C)$ is closed.

Definition: A continuous surjection is called perfect if it is closed and the preimages of points are compact.

Definition: A continuous function is called proper if the preimages of compact sets are compact.

Lemma 4 (without proof): A perfect map is proper.

Proposition: Let $F\rightarrow E\xrightarrow{\pi}B$ be a fiber bundle with compact fiber $F$. Then $\pi$ is a proper map.
Proof: $B$ is covered by trivialized open sets $\{U_i\}$. Since $F$ is compact and $\pi_i:\pi^{-1}(U_i)\to U_i$ is equal to the projection $U_i\times F\to U_i$ up to homeomorphism, it is a closed map by Lemma 1. By lemma 3 the entire map $\pi$ is closed. Since fibers are compact, $\pi$ is also a perfect mapping, hence it is proper by Lemma 4.

I decided to answer this question because it seems that there is a plethora of proofs, none of which is the one I think of first!

Note: I assume that compact means "compact and Hausdorff", which is the way I was taught at Warsaw University. This implies normality.

Claim: There is a finite cover $$U_{i}$$ of $$B$$ such that $$\pi$$ is trivial over some open neighbourhood $$V_{i}$$ of the closure of $$U_{i}$$, ie. $$\bar{U_{i}} \subseteq V_{i}$$.

Proof of the claim: By compactness it's enough to find a neigbourhood with that property of every point in $$B$$, so let $$b \in B$$. Let $$V_{b}$$ be an open neighbourhood of $$b$$. By normality, there are open sets $$\{ b \} \subseteq U_{b}, (B \setminus V_{b}) \subseteq W_{b}, U_{b} \cap W_{b} = \emptyset$$, because $$\{ b \}, (B \setminus V_{B})$$ are closed and disjoint. Now $$\bar{U_{b}} \subseteq V_{b}$$, since $$\bar{U_{b}} \subseteq (B \setminus W_{b}) \subseteq (B \setminus (B \setminus V_{b})) \subseteq V_{b}$$, so $$U_{b}$$ is the required neighbourhood.

This is a basic property of (locally) compact spaces which makes precise the statement that in such spaces "neighbourhoods of points can be made arbitrarily small". The theorem you're after follows easily.

Let $$U_{i}$$ be a cover of $$B$$ as above. Since $$\bar{U_{i}} \subseteq V_{i}$$, the subsets $$\pi^{-1}(\bar{U_{i}}) \simeq \bar{U_{i}} \times F$$ are compact and their union is $$E$$ by construction. So $$E$$ is the union of a finite number of compact subsets and thus is compact itself.

• Does compactness alone imply normality? – Jason DeVito Mar 8 '13 at 17:58
• You are using regularity, but the space is not assumed to be regular or Hausdorff, it is rather as general as possible. – Stefan Hamcke Mar 8 '13 at 21:12
• I am sorry, but I believe in many areas of mathemathics it is customary to call a space "compact" to mean compact and Hausdorff, which implies normality. This convention is so widespread that the term "quasi-compact" is introduced to mean compact, not necessarily Hausdorff space. Anyway, I will edit my answer to make it explicit that I assume the spaces involved to be Hausdorff. – Piotr Pstrągowski Mar 9 '13 at 13:15
• @Piotr: I see. Well +1 from me - I work with (Hausdorff!) manifolds, so this proof works great for me, even if it's not the most general. – Jason DeVito Mar 9 '13 at 14:37
• I guess Hausdorff'ness then follows from the fact that every point has a closed Hausdorff neighborhood, right? What if $F,B$ are not compact? Does Hausdorff'ness of $E$ still follow? – Stefan Hamcke Mar 9 '13 at 16:46