Timeline for A question about local connectedness
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2010 at 4:28 | comment | added | Ian Agol | See e.g. Henon attractor for a picture. en.wikipedia.org/wiki/Henon_attractor | |
| Aug 24, 2010 at 0:14 | comment | added | Victor Protsak | By the way, I don't think that a continuous image of a solenoid can be a pseudo-arc: while a solenoid isn't locally connected, it does contain a closed arc through every point, which would have to contract to a point, since a pseudo-arc doesn't contain a nondegenerate pathwise connected subcontinuum. | |
| Aug 23, 2010 at 22:51 | comment | added | Victor Protsak | Thank you, that does explain it. The fallacy in my objection was a silly circular reasoning: the closed neighborhood would be homeomorphic to the whole pseudo-arc iff it is a subcontinuum, i.e. connected (which is what we are trying to determine). Meanwhile, I found a much more elementary example and updated my answer. | |
| Aug 23, 2010 at 22:29 | comment | added | Bill Johnson | No, because e.g. a locally connected continuum is arcwise connected. (Continuum is important here; there are connected and locally connected metric spaces that do not contain any arc.) There are strong uniqueness properties of the pseudo-arc. I don't remember what they are but suspect that they imply that your example is a pseudo-arc. | |
| Aug 23, 2010 at 22:04 | comment | added | Victor Protsak | Bill, pseudo-arc was my first thought, but since it's homeomorphic to its nondegenerate subcontinua, doesn't this mean that its closed metric neighborhoods are connected and, consequently, that it is locally connected at each point? | |
| Aug 23, 2010 at 21:36 | history | answered | Bill Johnson | CC BY-SA 2.5 |