Timeline for In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2020 at 22:55 | comment | added | Mikhail Ostrovskii | @NikhilSahoo You are welcome. One can extend this argument to some classes of nonseparable spaces. It can happen that the general nonseparable case is difficult. | |
| Jul 24, 2020 at 22:52 | history | edited | Mikhail Ostrovskii | CC BY-SA 4.0 |
added 34 characters in body
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| Jul 24, 2020 at 10:16 | comment | added | Jochen Wengenroth | This is of course a very good answer -- although it does not answer the question as stated. Perhaps it should not have been accepted so quickly. | |
| Jul 24, 2020 at 6:34 | comment | added | Nikhil Sahoo | Thanks! For the past two days, I've been grasping at ideas regarding subsequences of bounded sequences, so it feels affirming to know that I was on the right track. Even if there may still be some non-separable example out there, I'm quite happy with this answer to the negative in the separable case, since sequential conditions on topological spaces tend not to work as nicely without some sort of countability condition. | |
| Jul 24, 2020 at 6:26 | vote | accept | Nikhil Sahoo | ||
| Jul 26, 2020 at 4:34 | |||||
| Jul 24, 2020 at 5:52 | history | answered | Mikhail Ostrovskii | CC BY-SA 4.0 |