Let H be a separable and infinitedimensional Hilbert space and let B be the closed ball of H having unit radius, whose center is at the origin h of H. Suppose one would like to know how much of B can be "filled up" by any of its compact subsetssince B itself (although closed and bounded) is not compact. Let E be the set of all positive real numbers z for which there exists a compact subset C of B such that all points of B lie at a distance from C (in the metric of H) which is not greater than z. The greatest lower bound of E would be a measure of this "filling up". My question iswhat is this greatest lower bound? I believe that it is 1 but cannot prove it. Clearly 1 belongs to E since we can take for C any compact subset of B that contains h. I can prove that no positive real number less than onehalf the square root of 2 belongs to E. But this is as far as I have been able to get. If 1 is the right answer, it would show that no compact subset of B can "fill up" any more of B than the set containing only the point h.
Instead of using compact subsets one can just use finite subsets. Now, consider any finite subset of $B$: it is contained in a finite dimensional subspace $V$. There is a unit vector perpendicular to $V$, and such vector has distance at least $1$ from the given finite subset.
Assume $C \subset H$ is compact such that each point of the unit disk has distance at most $1\epsilon$ from $C$ for some $\epsilon > 0$. There are finitely many points $p_1,...p_n$ such that the balls with radius $\frac{\epsilon}{4}$ around the $p_i$ cover all of $C$. Now pick an orthonormal basis $e_i$ of your Hilbert space and expand the $p_i$ in this basis, say $p_i = \sum\limits_j a_{ij} e_j$. Since there are only finitely many $p_i$, there must be an index $k$ such that $\left\vert a_{ij} \right\vert < \frac{\epsilon}{4}$ for all $i$ and all $j \geq k$. Then $e_k$ has at least distance $1\frac{\epsilon}{4}$ from all the $p_i$. Since each point of $C$ has distance at most $\frac{\epsilon}{4}$ from one of the $p_i$, no point of $C$ can be closer than $1\frac{\epsilon}{2}$ to $e_k$, contradicting the original choice of $\epsilon$.
What you are describing is exactly the ball measure of noncompactness of the closed bounded set $B$: $$\alpha(B):=\inf \{r>0 \, : \exists F\subset B, \mathrm {\, F\, finite\, ,s.t.} B\subset \cup_{x\in F} B(x,r) \}\, .$$ It can be viewed as the pointset distance $\inf_{F\in \mathcal{F}}d(B,F)$ between $B$ (as a point of the metric space $\mathcal{H}$ of all closed bounded subsets of $H$ endowed with the Hausdorff distance), and the set $\mathcal{F}\subset \mathcal{H}$ of all finite subsets of $H$. Since $\mathcal{F}$ is dense w.r.to the Hausdorff distance in the set $\mathcal{K}\subset \mathcal{H}$ of all compact subsets, one may equivalently use $\mathcal{K}$ instead of $\mathcal{F}$.
That the ball measure of the unit ball in a infinite dimensional Banach space is 1 and not less, follows immediately by a simple rescaling argument. If you can cover $B$ with $N$ balls of radius $\theta < 1$, then you can also cover it by $N^2$ balls of radius $\theta^2$, and by $N^k$ balls of radius $\theta^k$, for any $k$, which in turn implies that it is relatively compact, contradiction.

$\begingroup$ To Fabian and Pietrothanks alot. These are very neat answers indeed. $\endgroup$ – Garabed Gulbenkian Dec 13 '11 at 23:40