A question about tiling Hilbert Space

Question

Let H be an infinite dimensional and separable Hilbert Space. Let e be a positive real number-which can be arbitrarily small. Does there exist a denumerably infinite set S of pairwise disjoint and pairwise congruent subsets of H such that (1) the union of S is H and (2) each set which is an element of S has a diameter not greater than e? Although the sets belonging to S must be bounded they do not have to be open or closed in H. Note that if H were a finite dimensional Euclidean space, my question would have an affirmative answer-since the sets belonging to S could be multi-dimensional semi-closed intervals.........The motivation for asking this question is to obtain a clearer understanding of the problems which arise when one attempts to define a measure for infinite dimensional spaces that is (to the greatest extent possible) a generalization or extension of n-dimensional Lebesgue measure.

Answer 1

Yes, there are such tilings, but I don't know of any "nice" ones. The example below makes use of the axiom of choice, so you can guess how horrible it is. :)

Take any dense countable subgroup $J$ of $H$ - say, the subgroup of vectors with rational coordinates, such that finitely many of those are nonzero. Now for every coset $x + J$ there is a representative of norm $\le \varepsilon$. Choose such a representative from every coset, and let's call $A$ the set of these. Now $A$ is bounded by construction, $A + j, j \in J$ are disjoint, and their union is the whole $H$ (since every coset contains exactly one representative in $A$).