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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?

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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 0:02

1 Answer 1

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.)

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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 0:07
    
@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 3:31
    
@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 3:58
    
@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 4:01
    
@ 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 4:14

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