Timeline for Iterated function system on the plane
Current License: CC BY-SA 3.0
9 events
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| Jul 11, 2013 at 21:53 | history | edited | MichaelNgelo | CC BY-SA 3.0 |
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| Jul 11, 2013 at 11:40 | comment | added | Emil Jeřábek | @MichaelNgelo: I’d suggest you correct the post, rather than delete it. This is a simple construction showing that the answer is yes (even using just homotheties) if $\max\{r_1+r_2,r_1+r_3,r_2+r_3\}\le1$, which complements Pablo Shmerkin’s answer. | |
| Jul 11, 2013 at 3:47 | comment | added | MichaelNgelo | @John, the self-similar pieces being disjoint when $r_i$'s are small is probably not a problem. The problem arises when $r_i+r_j$ are too large, as shown in other comments and Pablo Shmerkin's answer. I am sorry for my wrong answer. Should I simply remove this post? | |
| Jul 10, 2013 at 18:26 | comment | added | john mangual | We were told $r_1^2 +r_2^2 + r_3^2 < 1$ so $r_1, r_2, r_3 < 1$. If all three are equal, they start being disjoint at $r = \frac{1}{4}$. Yes, it doesn't have to be equilateral. | |
| Jul 10, 2013 at 17:02 | comment | added | Emil Jeřábek | This works (for any triangle) only if $R\le1$. | |
| Jul 10, 2013 at 16:59 | comment | added | Tapio Rajala | To see that this does not work take for instance $r_1=r_2=r_3 > \frac12$. | |
| Jul 10, 2013 at 16:51 | comment | added | john mangual | Solving for $K = f_1(K) \cup f_2(K) \cup f_3(K)$ this generalizes the Sierpinski triangle. | |
| Jul 10, 2013 at 16:45 | history | edited | MichaelNgelo | CC BY-SA 3.0 |
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| Jul 10, 2013 at 16:40 | history | answered | MichaelNgelo | CC BY-SA 3.0 |