Timeline for A contractible non-planar continuum
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2016 at 18:07 | comment | added | Anton Petrunin | @TarasBanakh you are right --- I overlooked this part. Should not be hard --- let me think. | |
| Jun 12, 2016 at 13:54 | comment | added | Taras Banakh | Anton, indeed, $\gamma_n$ is closed curve (but not necessarily a simple closed curve). The contradiction you derive is correct as soon as you prove that the component $\Omega_n$ (containing the point $\iota(0,0,0)$) is bounded. Why is it bounded? More precisely, at which place of the proof is it established? | |
| Jun 12, 2016 at 13:43 | comment | added | Anton Petrunin | @TarasBanakh $\gamma_n$ is closed (read carefully). You need to show that for fixed $\varepsilon>0$ the set $\Omega_n$ lies in $\varepsilon$-neighborhood of $\iota(\{0\}\times[-1,1]\times\{0\})$. If not you get a sequence of points $\iota(\pm\tfrac1n, a_n,0)$ which converges to a point outside of $\iota(\{0\}\times[-1,1]\times\{0\})$, a contradiction. | |
| Jun 12, 2016 at 13:38 | comment | added | Taras Banakh | Dear Anton thank you for your answer. I also thought this way (at first), but there are some problems in the above proof: first, the set $\gamma_n$ is not a closed curve. This can be fixed choosing an appropriate closed curve inside of $\gamma_n$. Next, the inclusion $\bigcap_{n>N}\Omega_n\subset \iota([-1,1]\times\{0\}\times\{0\})$ is not so clear: we should prove at first that the component $\Omega_n$ is bounded + check many other unpleasant details. So, the proof should be polished at least. (That is why I put this "obvious" question, realizing that it is not so obvious). | |
| Jun 12, 2016 at 13:26 | history | answered | Anton Petrunin | CC BY-SA 3.0 |