# compactness of Banach spaces

I'm stuck on the following problem:

a) Let X be a Banach space, $K \subset X$ non-empty and compact and $\epsilon > 0$. Show that $K^{\epsilon} := \{k+x | k \in K, \|x\| \leq \epsilon\}$ is not compact if X is infinite dimensional.

I was thinking that the proof should be similar to that showing that the unit sphere is not compact, using Riesz' lemma. But when trying to construct a sequence without bounded subsequence I don't seem to see why the elements of this sequence should be in $K^\epsilon$.

EDIT: I think I have solved this, and it was actually quite simple. $K^\epsilon$ contains a sphere of radius $\epsilon$ around some point $k \in K$, now take a sequence of unit vectors without a convergent subsequence (which exists because the unit sphere is not compact), and translate that to $K^\epsilon$. Still stuck on question b though.

b) Let $\{A_k\}_{k \in \mathbb{N}}$ be a family of compact subsets of X such that there exist $r_k > 0$ with $A_{k+1} \subset A_k^{r_k}$ and $\sum r_k < \infty$. Show that the closure of $\bigcup_{k=1}^{\infty} A_k$ is compact.

I'm not really sure how to start here...

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I tried to answer a similar question yesterday, and botched it. But, at the risk of embarrasing myself again ... I'm not sure this is true. Your set $K^\epsilon$ is $K + \epsilon B$, where $B$ is the unit ball. The unit ball is closed (I think) and $K$ is compact, and this implies that $K + \epsilon B$ is compact. –  bubba Oct 17 '12 at 5:46
Take K = $\{0\}$ and $\epsilon = 1$, then your implication is not true. –  III Oct 17 '12 at 6:39