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Theorem. Every Manifold is locally compact.

This is a problem in Spivak's Differential Geometry.

However, don't know how to prove it. It gives no hints and I don't know if there is so stupidly easy way or it's really complex.

I good example is the fact that Heine Borel Theorem, I would have no clue on how to prove it if I didn't see the proof.

So can someone give me hints. I suppose if it's local, then does this imply that it's homeomorphic to some bounded subset of a Euclidean Space?

Pete L. Clark
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simplicity
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    Hint: every point has a neighborhood homeomorphic to the open unit ball in $\mathbb R^n$. – GEdgar Jan 29 '12 at 23:56
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    I took the liberty of editing out your second question, which I really think you should post as a separate question since it has no direct relationship to your first question. (Also, the honest answer to "How do prove Invariance of Domain?" is "Look it up." It is certainly too hard for a nonexpert to prove as an exercise. And in fact if you google for -- invariance of domain, proof -- you will find plenty of proofs...and see that they are not so easy.) – Pete L. Clark Jan 30 '12 at 00:02

4 Answers4

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By definition, if $X$ is a manifold, then every point $x \in X$ admits an open neighborhood $U$ which is homeomorphic to $\mathbb{R}^n$ ($n$ is allowed to depend on $x$). Let $f: U \rightarrow \mathbb{R}^n$ be such a homeomorphism. Let $B$ be a closed ball of finite radius about $f(x)$ in $\mathbb{R}^n$. By Heine-Borel, $B$ is compact, hence so is its homeomorphic preimage $f^{-1}(B)$, which is therefore a compact neighborhood of $x$.

Almost the same argument shows that $X$ has even a neighborhood base of compact sets at every point, which for non-Hausdorff spaces, is a priori stronger than having a single compact neighborhood at any point. In my opinion "locally compact" should mean this stronger condition. (On the other hand, in my terminology, both "manifold" and "locally compact" include the Hausdorff condition.)

Pete L. Clark
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  • Thanks for that. I knew you had to use something like Heine Borel.On n depending x I sort of don't understand that. But, might ask it tomorrow. Thanks for that. – simplicity Jan 30 '12 at 00:07
  • @Pete Intuitively, the statement of the claim makes sense since by definition, a manifold is a second countable topological space which is locally homeomorphic to Euclidean space. Since Euclidean space has compact subsets by the Heine-Borel property and compactness is a topological property, i.e. preserved under homeomorphisms, we should be able to "pullback" compact subsets of Euclidean space into compact subsets of the manifold via local homeomorphisms. Of course,seeing a claim is reasonable is one thing,proving it is another. In this case,it's just straightforward point set topology. – Mathemagician1234 Jan 30 '12 at 03:31
  • @Mathemagician: Yes, I agree that the statement in question is very straightforward to prove. But the OP seems to be a beginner in this area and is asking for help with these straightforward things. I think it is more than fair for a student to ask for something basic to be spelled out in full detail, so I answered in that way. – Pete L. Clark Jan 30 '12 at 03:58
  • @simplicity: The point is for instance that the subspace set of points in $\mathbb{R}^3$ such that ($z = 0$) or ($z = 1$ and $x^2 + y^2 = 1)$ is a manifold, even though some points have neighborhoods homeomorphic to $\mathbb{R}^2$ and some have neighborhoods homeomorphic to $\mathbb{R}$. Since the $n$ is unique for a given $x$ (this has something to do with invariance of domain!) and the function $x \mapsto n(x)$ is continuous into a discrete space, it is actually constant on the connected components of the manifold. In other words, $n$ only depends on $x$ in a very simple way. – Pete L. Clark Jan 30 '12 at 04:01
  • @ Pete I completely agree and agree with your assessment of the neophyte level of the OP. And I think it was totally appropriate for you to completely fill in the blanks for him or her. You seem like an excellent teacher from what I've seen of you on this site,as well as your writings on your homepage and this is just a further example of that. – Mathemagician1234 Jan 30 '12 at 04:14
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Manifolds are locally Euclidean, and Euclidean space is locally compact. Hence manifolds are locally compact.

pre-kidney
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Well, recall that manifolds locally look like euclidean space, by "look like", I mean locally homeomorphic to. Therefore, topological properties in the local sense should be translatable from the euclidean sense to the manifold sense. I shall write a proof keeping this intuition in mind.

Recall that a topological space is locally compact if each point $p$ in $M$ has a neighborhood $U$ that is contained in a compact set.

Proposition: If $M$ is a topological manifold, then $M$ is locally compact.

Proof: Let $p\in M$. Let $(U,\phi)$ be a chart about $p$ where $\phi:U\rightarrow \phi(U)$ is a homeomorphism and $\phi(U)$ is an open subset of $\mathbb{R}^n$ for some $n$. Let $r>0$ be a real number such that $B=B(\phi(p),r)\subseteq \phi(U)$. Shrinking $r>0$ if necessary, we may assume that $\overline{B}=\overline{B(\phi(p),r))} \subseteq \phi(U)$. Hence, $p\in \phi^{-1}(B)\subseteq \phi^{-1}(\overline{B})\subseteq U$. Therefore we are done.

Moe Khaled Bin-Lateef
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I do not think the above answers are completely right, since the "Hausdorff" condition in the definition of topological manifolds must be needed.

The key is to prove the following: If $V\subset U\subset X$, and X is Hausdorff, $\bar{V}_{U}$ is compact. Then $\bar{V}_{U}=\bar{V}$.

Proof: By definition, we only need to show that $\bar{V}\subset \bar{V}_U$. Suppose $x\in\bar{V}$, if $x\notin \bar{V}_U$. Since X is Hausdorff and $\bar{V}_{U}$ is compact, $\bar{V}_{U}$ is closed in X. So $ (\bar{V}_{U})^{c}$ is open and $x\in (\bar{V}_{U})^{c}$. Since $x\in \bar{V}$, we must have $V\cap (\bar{V}_{U})^{c}$ is not empty. It is impossible since $\bar{V}_{U}$ contains $V$.

Koobe
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  • Actually you don't need the Hausdorff condition. A manifold is "locally Euclidean", and locally Euclidean implies locally compact. This is even with a "non-Hausdorff manifold", as the line with two origins for example. – PatrickR Mar 22 '23 at 02:49
  • you are right, but I can't see how to prove locally Euclidean implies locally compact. And my definition for locally compact is that every point has a neiborhood whose closure is compact. All the above answers take the closure in the chart instead of in the manifold – Koobe Mar 31 '23 at 21:06
  • see definition of locally compact : https://en.wikipedia.org/wiki/Locally_compact_space – Koobe Mar 31 '23 at 21:21
  • I was using locally compact = condition (3) in wikipedia = this condition in pi-base. Your condition is condition (2) from wikipedia = this condition in pi-base. I think you are right: loc Euclidean implies condition (3) but probably not condition (2) (although the line with two origins does satisfy condition (2) and (3). FYI the search box in pi-base shows examples of (2) and not (3), and also of (3) and not (2). It would be good to find an example of locally Euclidean without cond (2). – PatrickR Mar 31 '23 at 22:59
  • This seems instructive. Examples of loc compact in sense (2) without being regular (which would be sense (4): here. For examples the Sierpinski space. – PatrickR Mar 31 '23 at 23:07
  • That is really helpful. Thank you – Koobe Mar 31 '23 at 23:47