# Is any closed ball non-compact in infinite dimensional space?

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?

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Yes. Take your sequence $\{x_n\}$ and consider the sequence $\{r x_n\}$. – David Mitra Dec 7 '12 at 14:24

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.