Timeline for Can you explicitly write $\mathbb{R}^2$ as a disjoint union of two totally path disconnected sets?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| Feb 25, 2013 at 6:53 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typo
|
| Feb 25, 2013 at 5:14 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
clearer introduction
|
| Feb 25, 2013 at 4:57 | comment | added | Włodzimierz Holsztyński | Images $f(I)$ of paths $f\rightarrow X$ in arbitrary metric spaces $X$ are the same as the connected and locally connected compact subsets of $X$ (Hahn-Mazurkiewicz Theorem). Each of them can be decomposed into a union of two smaller continua, if it contains more than one point. On the other hand the Knaster pseudo-arc $K$ is hereditarily indecomposable (all subcontinua of $K$ are homeomorphic to $K$, how nice!) hence it does not contain any image of a non-constant path. Probably I could use some of it to make the proof above a bit nicer. | |
| Feb 25, 2013 at 4:38 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
format
|
| Feb 24, 2013 at 21:01 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |