Is any closed ball non-compact in an 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?
 A: 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?
Yes. Consider the proof of Theorem 1.22 in Rudin's book, according to which Every locally compact topological vector space $X$ has finite dimension. If $X$ is a normed space and we consider the particular neighborhood  $V=B(0,1)$ then such a proof takes this simpler form. By rescaling and translating, we get the argument below.

Claim. Any closed ball is non-compact in an infinite dimensional space.

Proof: Let $X$ be a normed space and $B$ an arbitrary closed ball in $X$, say with center $a$ and radius $r$.
Assume that $B$ is compact.
As $\{B(x;\frac{r}{2})\}_{x\in B}$ is an open cover of $B$, there are $x_1,...,x_n\in B$ such that $$B\subset \bigcup_{i=1}^n B(x_i;\tfrac{r}{2}).$$
As
$$B(x_i;\tfrac{r}{2})= x_i-a+B(a;\tfrac{r}{2})=x_i-\tfrac{a}{2}+\tfrac{1}{2}B(a;r)$$
it follows that
$$B(a;r)\subset B\subset \bigcup_{i=1}^n \big(x_i-\tfrac{a}{2}+\tfrac{1}{2}B(a;r)\big)\subset Y+\tfrac{1}{2}B(a;r),$$
where $Y=\text{span}\{a,x_1,...,x_n\}$. Thus
\begin{align*}
B(a;r)\subset Y+\tfrac{1}{2}[Y+\tfrac{1}{2}B(a;r)]&=Y+\tfrac{1}{2^2} B(a;r)\\
&\subset Y+\tfrac{1}{2^2} [Y+\tfrac{1}{2}B(a;r)]=Y+\tfrac{1}{2^3}B(a;r).
\end{align*}
In general,
$$B(a;r)\subset Y+\tfrac{1}{2^n} B(a;r)=Y+B\big(a,\tfrac{r}{2^n}\big),\quad \forall\ n\in\mathbb N.$$
So, each $x\in B(a;r)$ has the form $x=y_n+x_n$ with $y_n\in Y$ and $x_n\in B\big (a;\tfrac{r}{2^n}\big)$. As $x_n\to a$, $y_n\to x-a$ and thus $x-a\in \overline{Y}=Y$. As $a\in Y$, it follows that $x=(x-a)+a\in Y$. This shows that $B(a;r)\subset Y$ which implies that $X\subset Y$ and thus
$$\dim X\leq\dim Y=n+1<\infty.$$
Therefore if $\dim X=\infty$ then $B$ is not compact. $\square$
A: 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.
