TITLE: Complete implies locally compact in length metric space?
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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 $[\langle 0,0 \rangle]_\sim$.