Timeline for Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections?
Current License: CC BY-SA 2.5
10 events
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| Jul 28, 2011 at 14:48 | vote | accept | Qiaochu Yuan | ||
| Nov 30, 2009 at 19:14 | comment | added | Andrew Stacey | @Qiaochu: I didn't think that it did (I think I missed an edit somewhere as well which is why I thought of it). It'd be nice to find an "if-and-only-if" for your question, or at least some interesting "wacky" examples. Is your interest focussed on already-niceish spaces (eg Hausdorff) or are all topological spaces up for game? | |
| Nov 30, 2009 at 16:50 | comment | added | Qiaochu Yuan | Editing does not, in fact, trigger a notification. I like that answer! | |
| Nov 22, 2009 at 21:49 | comment | added | Gian Maria Dall'Ara | Very nice proof! Now, even if it is not a characterisation, it's much more satisfying. | |
| Nov 22, 2009 at 21:13 | comment | added | Andrew Stacey | Note that I got rid of both metrisability and sigma compactness now. | |
| Nov 22, 2009 at 21:12 | history | edited | Andrew Stacey | CC BY-SA 2.5 |
Much simpler proof with much weaker hypotheses.
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| Nov 20, 2009 at 13:51 | comment | added | Andrew Stacey | I suspect that sigma compactness is too much as well, or at least with paracompactness can be replaced by a sort of local sigma compactness. I'll be interested to learn if there's a simple description of all the topological spaces that have this property. However, for this answer I just wanted something that would work and wasn't too restrictive. | |
| Nov 20, 2009 at 13:30 | comment | added | Gian Maria Dall'Ara | Sigma compactness is a natural condition, which is useful even in measure theory since it ensures the regularity of measures given by Riesz representation theorem. | |
| Nov 20, 2009 at 13:30 | comment | added | Gian Maria Dall'Ara | Metrizability seems too much. After all, what you use is sigma compactness of every open set and the existence of the function going to zero fast enough. If K_n is the exhaustion of U by compact sets than a positive map whose value is b_n on Cl(K_n\K_n-1) and 0 outside Int(K_n+1\K_n-2) should be enough. So I think the right condition, at least for your argument to work is the existence of bump functions (loc comp hausdorff) and sigma compactness of open sets. | |
| Nov 20, 2009 at 12:01 | history | answered | Andrew Stacey | CC BY-SA 2.5 |