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Aug 2, 2016 at 0:22 answer added D.S. Lipham timeline score: 8
Oct 22, 2011 at 4:03 answer added Sergey Melikhov timeline score: 4
Aug 1, 2010 at 22:27 comment added Gerald Edgar Hilbert space is homeomorphic to $\mathbb{R}^\mathbb{N}$ ... so you could ask this question there instead if it helps.
Aug 1, 2010 at 14:46 answer added BS. timeline score: 3
Jun 7, 2010 at 15:14 comment added Garabed Gulbenkian The theorem I stated above about infinite connected closed sets holds in R^n because all closed subsets of R^n are countable unions of compact sets. But not all closed subsets of an infinite dimensional and separable Hilbert space ,H, have this property. Consider-for example-a closed ball in H. This is why the theorem fails to hold in H. This failure, in turn, leaves open several questions about the properties of closed and conmnected subsets that might possibly exist in H.
May 31, 2010 at 19:49 comment added Garabed Gulbenkian To Pietro Majer: One analog is-If E(M,p,e) denotes the set of all points of M lying within any positive distance e of any point p belonging to an infinite closed connected subset M of R^n, then E(M,p,e) contains an infinite connected subset. This theorem does not hold in an infinite dimensional and separable Hilbert space.
May 19, 2010 at 1:36 comment added François G. Dorais I did remember correctly, see Theorem 4.17 and the following remark in Kechris's Classical Descriptive Set Theory.
May 18, 2010 at 21:57 comment added François G. Dorais IIRC the universal properties of $\ell_2$ (in the realm of Polish spaces) guarantee that if there is a Polish space with these properties then there is one which is a closed subspace of $\ell_2$.
May 18, 2010 at 20:26 comment added Pietro Majer ..and lines certainly have closed intervals in them ;) Garabed: what's the analog result for H= R^n ?
May 18, 2010 at 20:15 comment added some guy on the street Hilbert space certainly has lines in it, if that's what you mean.
May 18, 2010 at 20:05 history asked Garabed Gulbenkian CC BY-SA 2.5