0

Let $S$ be a bounded sigma-compact subset of a separable infinite dimensional Hilbert space. Can we always find an increasing sequence of compact sets $K_n$ such that $S=\cup_nK_n$ and for any compact subset K of S there exists a natural number $N$ such that K is contained in the union of the first N of the $K_n$?

Arctic Char
  • 16,007
kers_bun
  • 35
  • 2
  • 1
    A first-countable Hausdorff space with this property is necessarily locally-compact. Thus it must fail for any infinite-dimensional Hilbert space (we can take $S$ to be the unit ball of a countably infinite-dimensional linear subspace, for instance). – Tyrone Nov 19 '20 at 23:06
  • Oh you are right! Thanks very much for your answer, it is all clear now. – kers_bun Nov 19 '20 at 23:16
  • The space itself doesn't have this property but any $\sigma$-compact sub-set will. – s.harp Nov 20 '20 at 12:27
  • Could you please elaborate on this? I would be interested to see the proof. – kers_bun Nov 20 '20 at 12:50
  • Actually I was mistaken - the proof I had in mind used local compactness. The argument described by Tyrone is correct: A countably infinite dimensional subspace is $\sigma$-compact, if the statement is true the countable infinite dimensional subspace must then be hemi-compact and as such locally compact / finite dimensional, a contradiction. – s.harp Nov 20 '20 at 13:44

0 Answers0