Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 5

up vote 4 down vote accepted

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

share|improve this answer
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 $n \ne m$).

share|improve this answer

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.)

share|improve this answer
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.

share|improve this answer

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 $1$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.