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.
Best Answer
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.