Banach Spaces: Totally Bounded Subsets

Question

As an easy consequence of Riesz' lemma it is known that infinite dimensional Banach spaces possess bounded subsets that fail to be totally bounded. On the other hand in finite dimensional Banach spaces any bounded subset happens to be totally bounded as well. So the question arises wether totally bounded subsets always sit within finite dimensional spaces, or put in equivalent form:

Is there an infinite dimensional space which possesses a totally bounded subset not lying entirely in a finite dimensional subspace?

Answer 1

In fact, any infinite-dimensional Banach space $X$ has a compact subset that does not lie in any finite-dimensional linear subspace. Namely, define a sequence $x_n$ inductively such that $x_{n+1}$ is not in the linear span of $\{x_1, \ldots, x_n\}$ but $\|x_{n+1}\| < 1/n$. Then $\{0\} \cup \{x_1, x_2, \ldots\}$ is your set.

Answer 2

Sure: the image of the unit ball in $\ell^2$ under the map that sends the standard basis element $e_n$ to $e_n/n$ (with $\ge 1$) is Hilbert-Schmidt, so a compact operator, etc. Indeed, any sequence going to $0$ replacing $1/n$ produces (pre-) compact image.