Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is known that the closed unit ball $\overline{B_1(0)}$ in a normed space $X$ is compact if and only if $\dim X < \infty$. In particular, the $\overline{B_1(0)}$ is not compact if $\dim X = \infty$. The proof of this involves finding a sequence $\{ x_n \}_{n\in\mathbb{N}} $ with $||x_n|| = 1 $ for all $n \in \mathbb{N}$ such that $||x_n - x_m|| > \frac{1}{2}$ for all $n \neq m$. Then this sequence is a bounded sequence that is not a Cauchy-sequence, so it does not have a converging subsequence, so $X$ is not (sequentially) compact. The construction then goes with Riesz's lemma, by subsequently finding points with norm 1 that have distance greater than $\frac{1}{2}$ to the subspace generated by the preceeding points.

Now my question is, is any closed ball $\overline{B_r(0)}$ with $r>0$ non-compact in an infinite-dimensional space? It seems very intuitive, since topologically the balls are all diffeomorphic, and it would seem unlikely that for $r>1$ the larger ball is compact while $\overline{B_1(0)}$ is not. However, when I look at the proof, I notice that Riesz' lemma really only works for norm 1 and not for some arbitary norm. Is there any way to adapt this lemma or make use of a different construction in order to say something about all the closed balls?

share|cite|improve this question
Yes. Take your sequence $\{x_n\}$ and consider the sequence $\{r x_n\}$. – David Mitra Dec 7 '12 at 14:24
up vote 6 down vote accepted

If $x$ is a vector in your infinite dimensional space and $\alpha\neq 0$, then the functions $x\mapsto v+x$ and $x\mapsto \alpha x$ are homeomorphisms of the space onto itself. It follows that all closed balls are homeomorphic and therefore fail to be compact.

share|cite|improve this answer
Ok, thanks, I did not see that compactness is a topological property, so it is preserved under homeomorphisms. – Yiteng Dec 7 '12 at 14:47

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.