Timeline for Is the union of a compact and the relatively compact components of its complementary in a manifold compact?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2022 at 17:10 | vote | accept | Saúl RM | ||
| Dec 13, 2021 at 9:34 | comment | added | Pierre PC | Ah yes, I see. I would say the remaining components are rather uninteresting though. I would probably consider only the inverse image of the interesting component in $M/K$ or so. | |
| Dec 11, 2021 at 18:00 | comment | added | Aitor Iribar Lopez | @PierrePC I think you also need $M$ to have only finitely many connected components, if not to be connected itself. Example: $M=\mathbb N$ with the discrete topology is LCH and locally connected, but if $X=\{1\}$ then $Y = \mathbb N$ is not compact. | |
| Dec 11, 2021 at 9:48 | comment | added | Pierre PC | This is a wonderful proof! What do we need about $M$, it should be Hausdorff, locally compact, locally connected? This would rule out Jochen Wengenroth's example above. | |
| S Dec 9, 2021 at 4:20 | review | First answers | |||
| Dec 9, 2021 at 4:56 | |||||
| S Dec 9, 2021 at 4:20 | history | answered | Aitor Iribar Lopez | CC BY-SA 4.0 |