I am confused. The way I see it, in a complete metric space, closed balls of finite diameter are compact since they are complete and totally bounded. Consequently a complete metric space is locally compact. Why/how/does this fail in a length metric space?
The reason I am asking this is because in the Hopf-Rinow theorem for length spaces, the hypothesis are that a space needs to be complete AND locally compact (I would think complete implies locally compact by the above reasoning?)…
Thanks for the help.
Best Answer
No, a complete space need not be locally compact. A closed ball, although it is bounded, need not be totally bounded. The standard examples are infinite-dimensional spaces, such as an infinite-dimensional Hilbert space $H$. The closed ball $B(0,1)$ centered at the origin with radius 1 is not compact, because an orthonormal basis $\{e_n\}$ provides a sequence in $B(0,1)$ with no convergent subsequence (indeed, by the Pythagorean theorem we have $d(e_n, e_m) = \sqrt{2}$ for all $n \ne m$).