Timeline for Is it possible to separate two linked (geometric) circles in $\Bbb R^3$ by a set homeomorphic to the 2-sphere (with arbitrarily “bad” homeomorphism)?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2022 at 6:48 | vote | accept | Mikhail Patrakeev | ||
| Jul 26, 2022 at 19:14 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
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| Jul 26, 2022 at 19:11 | comment | added | Anton Petrunin | @MikhailPatrakeev I simplified the argument considerably. | |
| Jul 26, 2022 at 19:10 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
simlification
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| Jul 26, 2022 at 18:57 | history | edited | LSpice | CC BY-SA 4.0 |
`\mathbb{R^3}` -> `\mathbb{R}^3`, and typo
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| Jul 26, 2022 at 18:52 | comment | added | Anton Petrunin | @MikhailPatrakeev $\tilde S$ is compact, so it lies in a ball, say $B$. The complement $\mathbb{R}^3\setminus B$ is connected --- so the other connected component have to be in $B$. | |
| Jul 26, 2022 at 17:37 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
edited body
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| Jul 26, 2022 at 16:51 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
edited body
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| Jul 26, 2022 at 16:43 | history | answered | Anton Petrunin | CC BY-SA 4.0 |