So while I was solving some problems on differential geometry, I stumbled upon a problem which is to show that every manifold is locally compact. Now, there is a proof for it here, but I was thinking of another way to show this.

Let $M$ be a manifold and let $p$ be some point in $M$. Then there exists a open set $V$ such that $p\in V\cong\mathbb{R}^n$. Therefore, $V$ is metrizable an let $d:V\times V\to\mathbb{R}$ be a metric on $V$. Now choose $r\in\mathbb{R}$ small enough such that the closed ball $B=\{q\in M\mid d(p,q)\leq r\}\subset V$. Since $B$ is compact and the choice of $p$ is arbitrary then $M$ is locally compact.

Is that ok?

(EDIT) I'm using the fact that since $V\cong\mathbb{R}^n$ then we can use Heine-Borel on $V$. Is that correct?

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    $\begingroup$ How do you know that $B$ is compact? $\endgroup$ – Daniel Fischer Aug 18 '15 at 9:06
  • $\begingroup$ Is there a counter example for this? @DanielFischer $\endgroup$ – Adel Saleh Aug 18 '15 at 9:13
  • $\begingroup$ Hmmn. From Heine–Borel, a closed unit ball of any finite-dimensional normed vector space is definitely compact. I don't think that this is true for infinite dimensions. It then follows by logical implication that a normed vector space is finite-dimensional if and only if its closed unit ball is compact. $\endgroup$ – Autolatry Aug 18 '15 at 9:41
  • $\begingroup$ But $\mathbb{R}^n$ is finite dimensional and homeomorphic to $V$. Does that make $V$ finite-dimensional? $\endgroup$ – Adel Saleh Aug 18 '15 at 9:44
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    $\begingroup$ @Autolatry: That's true. A normed vector space is finite-dimensional iff it's locally compact or, equivalently, if the closed unit ball is compact. See terrytao.wordpress.com/2011/05/24/… $\endgroup$ – Stefan Hamcke Aug 18 '15 at 9:46

Your proof is wrong. You cannot conclude that $B$ is compact. Let $X$ be an infinite set and equip it with the discrete metric $d(x,y) = 1$ if $x \neq y$ and $0$ otherwise. $X$ is not compact, since the cover $\{x\}_{x \in X}$ admits no finite subcover. However, $X = \{y \in X: d(x,y) \le 2\}$ for any $x \in X$. In particular, you cannot conclude that your set $B$ above is compact.

  • $\begingroup$ Is there a way to use the metrizability of $V$ in the proof above to conclude that $M$ is locally compact? $\endgroup$ – Adel Saleh Aug 18 '15 at 10:00
  • $\begingroup$ In a sense, yes -- the metric on $V$ can be pushed forward to $\mathbb{R}^n$ by your homeomorphism where it generates the standard topology on $\mathbb{R}^n$, so it satisfies the Heine-Borel theorem. However, arguing in this way is essentially the linked proof anyway. The point I was trying to make is that it is not enough that you have a closed ball in a metrizable space. $\endgroup$ – guest Aug 18 '15 at 10:04
  • $\begingroup$ Actually, if we suppose that $V$ is a topological space homeomrphic to $\mathbb{R}^n$, and $B$ is a closed and bounded ball in $V$, then it's image by the hemeomorphism $f:V\to\mathbb{R}$, call it $f(V)$ is also closed and bounded in $\mathbb{R}^n$, hence is compact by Heine-Borel. Therefore, $B$ is compact since $B=f^{-1}(B)$ (f is a continuous bijection). $\endgroup$ – Adel Saleh Aug 18 '15 at 12:08
  • $\begingroup$ "Bounded" refers to a metric. It is not a topological property, and it is not invariant under continuous mappings. You are using the fact that the metric you picked up generates the standard topology on $\mathbb{R}^n$. $\endgroup$ – guest Aug 18 '15 at 12:13

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