Timeline for Inserting an open and simply-connected set between a compact set and an open set
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 15, 2015 at 22:17 | answer | added | Alex Zorn | timeline score: 0 | |
| Feb 14, 2015 at 18:49 | history | edited | smyrlis | CC BY-SA 3.0 |
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| Feb 11, 2015 at 23:10 | history | edited | smyrlis | CC BY-SA 3.0 |
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| Feb 11, 2015 at 14:34 | answer | added | Paul Fabel | timeline score: 5 | |
| Feb 11, 2015 at 14:24 | history | edited | smyrlis | CC BY-SA 3.0 |
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| Feb 11, 2015 at 14:14 | comment | added | smyrlis | $K$ is compact with connected complement in $\mathbb C$. | |
| Feb 11, 2015 at 13:48 | comment | added | Emil Jeřábek | What is the paper’s definition of “simply connected”? As demonstrated in Gabriel Drumond-Cole’s answer, the “more generally” part is not actually more general if one defines simply connected by null-homotopy of loops, as this does not imply connectedness of the complement of $K$. Perhaps the authors intend the latter as their working definition? | |
| Feb 11, 2015 at 11:43 | answer | added | Gabriel C. Drummond-Cole | timeline score: 7 | |
| Feb 11, 2015 at 11:12 | comment | added | Loïc Teyssier | @smyrlis: my apologizies, bad intuition ;) | |
| Feb 11, 2015 at 10:16 | comment | added | Francesco Polizzi | In fact, $K$ is retract of some (open) neighborhood if and only if it is weak locally contractible, see my answer below. | |
| Feb 11, 2015 at 10:11 | answer | added | Francesco Polizzi | timeline score: 7 | |
| Feb 11, 2015 at 9:43 | comment | added | smyrlis | If the boundary of $K$ is sufficiently smooth, then the tubular neighbourhood works. But when $K$ has a non-smooth boundary, one can construct a counterexample, where for every $\varepsilon>0$, the corresponding tubular neighbourhood is not simply connected. | |
| Feb 11, 2015 at 8:48 | comment | added | Francesco Polizzi | If $K$ is sufficiently smooth this is a consequence of the tubular neighborhood theorem (because $K$ is a retract of a sufficiently small $\epsilon$-neighborhood). | |
| Feb 11, 2015 at 8:21 | comment | added | Loïc Teyssier | Have you tried considering $\varepsilon$-neighborhoods of $K$ with $\varepsilon$ small enough ? I bet such neighborhoods are simply connected if $K$ is (when $\varepsilon$ is small enough). | |
| Feb 11, 2015 at 8:06 | history | asked | smyrlis | CC BY-SA 3.0 |