4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The unit ball is bounded by definition. Maybe you mean totally bounded in the non-Banach-space context? $\endgroup$ – Tomek Kania Aug 26 '14 at 21:03
  • $\begingroup$ @TomekKania: There was a typo in the question: Is there a totally bounded subset that... $\endgroup$ – C-Star-W-Star Aug 26 '14 at 21:10
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.