To prove that A normed space is locally compact if and only if its finite dimensional, I need to prove a lemma: A normed space is locally compact if and only if its closed unit ball is compact.

One way implication seems to be easy i.e., if the closed unit ball is compact then normed space is locally compact. But I'm still not very clear. How to prove the lemma?

As I understand (one of) the definition(s) of a locally compact metric space is:A metric space (X,d) is said to be locally compact if every x belongs to some open set A such that A is compact.

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    $\begingroup$ locally compact means that for any $x$ there is a compact set $B$ and an open set $A$ such that $x\in A\subset B$. It is also equivalent to ask for the existence of an open set $A\ni x$ s.t. $\bar A$ is compact - but the definition you wrote is not correct. $\endgroup$
    – Ilya
    Mar 21 '12 at 10:11
  • $\begingroup$ Hint: As $\cdot \lambda: X \to X$ is a homeomorphism for $\lambda \ne 0$, a subset $A$ of $X$ is compact iff $\lambda A$ is. $\endgroup$
    – martini
    Mar 21 '12 at 10:13
  • $\begingroup$ @Ilya Ah ok. so A need not be compact but A should belong to a compact set. right. thanks! $\endgroup$
    – Swair
    Mar 21 '12 at 10:15
  • $\begingroup$ Let $\overline B(x,d)$ denote the closed ball of diameter $d$ centered at $x$. Can you show that $\overline B(x,d)$ is compact $\Leftrightarrow$ $\overline B(0,1)$ is compact? $\endgroup$ Mar 21 '12 at 10:26
  • $\begingroup$ @MartinSleziak yes since, mapping x->ax+d is homeomorphism, $\overline B(0,1)$ is compact $\Rightarrow \overline B(x,d)$ is compact. And since $B(x,d) \in \overline B(x,d)$ in turn proves that the space is compact.right? $\endgroup$
    – Swair
    Mar 21 '12 at 10:34

The family of all balls is a base for the topology. Fix a point $x$. By the definition of local compactness, there exists $V$ open with $x\in V$ and $\overline V$ compact. Then there is a ball $B_\delta(x)\subset V$. The closure of this ball is in $\overline V$, so it is compact. But then $B_1(0)=\frac1\delta\,B_\delta(x)-x$ is compact, too.


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