# Every manifold is locally compact?

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?

• Hint: every point has a neighborhood homeomorphic to the open unit ball in $\mathbb R^n$. – GEdgar Jan 29 '12 at 23:56
• 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

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.)
• @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