Timeline for Inserting an open and simply-connected set between a compact set and an open set
Current License: CC BY-SA 3.0
14 events
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| Feb 11, 2015 at 14:43 | comment | added | Francesco Polizzi | @Gabriel: ah, yes! Now I was thinking about the existence of a simply connected neighborhood that retracts over the space, that of course is a much stronger requirement than the bare existence of a simply connected neighborhood. | |
| Feb 11, 2015 at 14:35 | comment | added | Gabriel C. Drummond-Cole | @Francesco, sure, of course it's not locally connected, but I think it's essentially obvious that it's not a counterexample to the question. | |
| Feb 11, 2015 at 14:31 | comment | added | Francesco Polizzi | @Gabriel: given a point in the limit vertical line (which belongs to the compact comb), all its neighborhoods are not connected, in particular the comb is contractible but not locally contractible. So I suspect that at least the comb is not a retract of an open neighborhood by Hatcher's theorem quoted in my answer (actually, one should actuallty check that it is not weakly locally contractible). | |
| Feb 11, 2015 at 14:29 | history | edited | Gabriel C. Drummond-Cole | CC BY-SA 3.0 |
edits in response to question edit
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| Feb 11, 2015 at 14:20 | comment | added | Gabriel C. Drummond-Cole | @Francesco I don't see why the comb has an open neighborhood with no simply connected open subspace. | |
| Feb 11, 2015 at 14:17 | comment | added | Francesco Polizzi | Anyway, taking the comb space instead of the Warsaw circle you have an example with a similar flavour and connected complement. en.wikipedia.org/wiki/Comb_space | |
| Feb 11, 2015 at 14:14 | comment | added | Gabriel C. Drummond-Cole | Maybe you should add the connected complement requirement to the question. | |
| Feb 11, 2015 at 14:12 | comment | added | smyrlis | This example fails to have connected complement. | |
| Feb 11, 2015 at 14:10 | comment | added | Francesco Polizzi | This is called Warsaw circle. It is a standard example of 1-dimensional continuum whose homotopy groups are all trivial, but which is neither contractible nor locally contractible. en.wikipedia.org/wiki/Continuum_%28topology%29 | |
| Feb 11, 2015 at 14:05 | comment | added | smyrlis | In the paper, $K$ is required to be a compact subset of $\mathbb C$ with a connected complement, i.e., not even connected. But if we further assume that $K$ is connected, then $K$ becomes simply connected (in the standard algebro-topological sense). | |
| Feb 11, 2015 at 14:01 | comment | added | Gabriel C. Drummond-Cole | My definition of simply connected is something like "connected and any two paths with the same endpoints are homotopic via an endpoint-preserving homotopy." | |
| Feb 11, 2015 at 13:50 | comment | added | Paul Fabel | Smyrlis, what is your definition of `simply connected'? | |
| Feb 11, 2015 at 13:46 | comment | added | smyrlis | But this $K$ is not simply connected! Neither its complement is connected. | |
| Feb 11, 2015 at 11:43 | history | answered | Gabriel C. Drummond-Cole | CC BY-SA 3.0 |