# 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.

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Closed balls of finite diameter are not compact in general. For example, the unit ball of an infinite dimensional Banach space (hence metric space in particular) is never compact. – Davide Giraudo Oct 4 '11 at 20:18
This seemed quite sufficiently answered already, and one answer was accepted, BUT a really simple answer was not included, so I've added it. – Michael Hardy Aug 3 '12 at 17:20

## 5 Answers

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 $[\langle 0,0 \rangle]_\sim$.

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Your example is very nice and I'm glad at least one answer addressed the length space issue. My favorite example is the real hyperbolic space of infinite dimension: take a Hilbert space and a form of signature $(\infty,1)$ and look at the set of vectors of norm $-1$. This gives an infinite dimensional complete CAT(-1)-space which is therefore also geodesic. – t.b. Oct 4 '11 at 23:29
@t.b.: Maybe this is a stupid question, but isn't a Hilbert space already a length space? Surely the straight line segments are geodesics? – Nate Eldredge Oct 5 '11 at 0:26
@Nate: No, it was more a stupid comment of mine... Of course, Hilbert spaces are length metric spaces, as are all normed spaces. I was thinking of something else for some reason, sorry about that. But while I'm at it, Hilbert spaces are characterized by being the only Banach spaces satisfying the CAT condition (which can be expressed via a variant of the parallelogram law). Once again, sorry about this slip. – t.b. Oct 5 '11 at 0:43
Is there any significance of the $2^{-k}$? I mean wouldn`t $\bigcup [0,1] \times \{k\}$ work just as well? – wspin Sep 3 '13 at 11:41
@wspin: (Sorry to be so slow.) Yes, it would. The $2^{-k}$ is useful for showing that this space is a counterexample to something else, but it’s not needed here. – Brian M. Scott Sep 15 '13 at 6:06

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 $n \ne m$).

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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.)

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Yes. But it is locally compact. – George Lowther Oct 4 '11 at 21:12
@GeorgeLowther: indeed, it's what's called a "local counterexample", against a specific claim made in the proof (in a complete metric space a closed ball is totally bounded), not a "global counterexample" against the complete statement. – Henno Brandsma Oct 5 '11 at 8:58

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.

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Closed balls of finite diameter in metric spaces are not always totally bounded.

For example, suppose there are infinitely many points and the distance between every two distinct points is $1$, and consider the closed ball of radius $1$.

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There is substantially no difference between your new answer and your old one. I voted this down because this duplication is really not useful. – t.b. Sep 14 '12 at 7:39