Timeline for Is every closed subset of finite measure contained in an open subset of finite measure?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2019 at 21:56 | comment | added | YCor | The distance is uniformly bounded by $1$. Since the whole space has infinite measure, this is not "locally finite" in your sense. | |
| Aug 9, 2019 at 21:51 | comment | added | Michael | Hmm, it's bounded by what ? | |
| Aug 9, 2019 at 21:19 | comment | added | YCor | On bounded subsets? this is not standard... in the answer you accepted, the whole space is bounded, so it doesn't work. | |
| Aug 9, 2019 at 21:13 | comment | added | Michael | I mean finite on bounded subsets | |
| Aug 9, 2019 at 20:51 | vote | accept | Michael | ||
| Aug 9, 2019 at 13:13 | history | edited | YCor | CC BY-SA 4.0 |
changed to more precise title
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| Aug 9, 2019 at 7:57 | history | became hot network question | |||
| Aug 9, 2019 at 4:49 | comment | added | YCor | What do you call "locally finite"? This terminology is usually used in locally compact spaces, where it equivalently means (a) finite on compact subsets (b) each point has a neighborhood of finite measure. The terminology would rather naturally mean (b), but it's better specify. | |
| Aug 9, 2019 at 0:22 | answer | added | Gerald Edgar | timeline score: 8 | |
| Aug 8, 2019 at 23:50 | history | asked | Michael | CC BY-SA 4.0 |