Timeline for Space $X$ such that $X^\lambda\cong X$ for some $\lambda$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2015 at 8:21 | vote | accept | Dominic van der Zypen | ||
| Jun 12, 2015 at 22:18 | answer | added | Garrett Ervin | timeline score: 9 | |
| Jun 12, 2015 at 11:35 | answer | added | Adam Przeździecki | timeline score: 4 | |
| Jun 11, 2015 at 8:42 | comment | added | Garrett Ervin | Any such $\lambda$ must be finite. For if $\lambda$ is infinite and $X \cong X^{\lambda}$, then by splitting off the first factor we have $X \cong X \times X^{\lambda}$, and therefore $X \cong X^2$. | |
| Jun 11, 2015 at 6:41 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |