Timeline for Is every real number in [0,1] a product of three (or more) Cantor set's numbers?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2023 at 21:11 | vote | accept | Pietro Majer | ||
| Dec 24, 2023 at 5:54 | comment | added | Pietro Majer | Ops, sorry, I forgot to delete this naive comment. Yes, i realized immediately after, $1/3\le \lambda\le3$ it is exactly the condition to have the property $F+\lambda F=[0,(1+\lambda)]$ invariant for the set-contraction $F\mapsto \frac13F+\{0,\frac23\}$, therefore, to have the fixed point $C$ there! | |
| Dec 23, 2023 at 23:29 | comment | added | Zach Teitler | @PietroMajer Well, it doesn't work for $\lambda=0$. For $0<\lambda<1/3$ I don't think it works either: Let $C_1 = [0,1/3] \cup [2/3,1]$. For $\lambda < 1/3$ it seems to me that $C_1 + \lambda C_1 = [0,1/3+\lambda] \cup [2/3,1+\lambda]$, so it misses $(1/3+\lambda,2/3)$. | |
| Dec 23, 2023 at 18:27 | comment | added | Pietro Majer | Nice paper… Am I wrong or theorem 3.1 (stating that $$C+\lambda C=[0,1+\lambda]$$ holds for every $1/3\le \lambda \le3$) in fact is also true for all $\lambda\ge0$ ? | |
| Dec 22, 2023 at 15:47 | history | answered | Zach Teitler | CC BY-SA 4.0 |