Complete implies locally compact in length metric space? 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.
 A: 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 $n \ne m$).
A: Others have already addressed the underlying misunderstanding; here’s a specific example of a complete length space that isn’t locally compact.
Let $$X = \bigsqcup_{k\in\omega}[0,1+2^{-k}] = \bigcup_{k\in\omega}\left([0,1+2^{-k}]\times\{k\}\right).$$ Let $\langle 0,k \rangle \sim \langle 0,n \rangle$ and $\langle 1+2^{-k},k \rangle \sim \langle 1+2^{-n},n \rangle$ for all $k,n\in\omega$ and $x\sim x$ for all $x\in X$; $\sim$ is an equivalence relation on $X$. Let $Y=X/\sim$. Then $Y$ is a complete length space with the obvious metric, but it’s not locally compact. Neither ‘endpoint’ has a compact nbhd: if $0<\epsilon<1$, $\{[\langle \epsilon/2,k \rangle]_\sim:k\in\omega\}$ is an infinite closed discrete set in the $\epsilon$-ball about $[\langle 0,0 \rangle]_\sim$.
A: Look at a metric space in which there are infinitely many points and the distance between any two points is $1$.  Closed balls of radius $1$ are not totally bounded in that space.  (And not compact.)
A: Complete metric spaces do not have to be locally compact. Closed balls of finite diameter do not have to be totally bounded.
For example, all infinite dimensional Banach spaces are complete metric spaces, but are not locally compact. In fact, a Hausdorff topological vector space is locally compact if and only if it is finite dimensional. As another example, Baire space is completely metrizable, but is not locally compact.
