Timeline for A question about homeomorphic subsets of Hilbert space
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2010 at 15:02 | comment | added | Garabed Gulbenkian | Call a point set "uncrowded" if it is denumerably infinite and if there exists a positive real number e such that every distinct pair of its points are at a distance apart not less than e. I was trying to prove the following two statements which would also imply a "yes" answer to my question. (!) Every closed and non-compact subset of H contains an "uncrowded" subset. (2) If A,B are "uncrowded" subsets of H then there exists a homeomorphism of H onto itself that carries A onto B. But your proofs, for which I thank you, are much better. | |
| Sep 5, 2010 at 16:52 | comment | added | Sergei Ivanov | Thanks, I was thinking about the radial projections but wrote about the point. Fixed now. | |
| Sep 5, 2010 at 16:51 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
correction: separated radial projections, not the points themselves
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| Sep 5, 2010 at 16:49 | comment | added | Pietro Majer | A minor point: maybe to get this family of $\epsilon$-separated rays one should take the family $\{ p_i \}$ in the radial projection of $A$ on the unit sphere, rather than in $A$ itself (otherwise some $p_i$ could be multiple of each other). | |
| Sep 5, 2010 at 16:43 | comment | added | fedja | A pointless nitpick: two $p_i$'s can be on the same ray (after all $p$ and $2p$ are $\varepsilon$-separated...), so the separation condition for rays doesn't hold automatically. It is trivial to fix though: the radial projection of $A$ to the unit sphere is also not compact. | |
| Sep 5, 2010 at 16:23 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |