Timeline for Mid point free sets
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2015 at 19:51 | comment | added | Ashutosh | @Mirko If X is a Borel and has positive measure, then it has a similar copy of every finite set of reals. You can prove this using Lebesgue density theorem. | |
| Apr 16, 2015 at 22:23 | comment | added | Mirko | @Ashutosh Thank you for the reference to Rado's result (nice proof). I wonder if there is a measurable non-null mid-point free set (I guess not, but I have no proof). There is a closed uncountable one (measure zero), follow the usual construction of the Cantor set but remove open intervals slightly bigger that the middle thirds. A paper by Jones about Hamel bases seems related (perhaps not directly). | |
| Apr 10, 2015 at 8:10 | comment | added | bof | In the literature, "mid point free sets" have been called "non-averaging sets". | |
| Apr 10, 2015 at 3:41 | comment | added | Mirko | Yes, under the additional assumption that $X$ is measurable. I posted an answer that has a gap, and my proof does not seem to work when $X$ is non-measurable with positive outer measure, but it does work when $X$ is measurable with positive measure. (The case when $X$ has measure zero is trivial, since then we could take $Y$ to be the empty set.) | |
| Apr 10, 2015 at 2:53 | answer | added | Mirko | timeline score: 0 | |
| Apr 10, 2015 at 0:25 | comment | added | Mirko | First thoughts. For $\varepsilon>0$ call $X$ $\varepsilon$-free if there are no $x,y,z\in X$ with $x+\varepsilon=y$ and $y+\varepsilon=z$. Given $\varepsilon$, is there $Z$ of inner measure $0$ such that $X\setminus Z$ is $\varepsilon$-free? (Note that it could conceivably happen that $m^*(X)=m^*(Z)=m^*(X\setminus Z$), see this answer.) Could we also get that $X\setminus Z$ is $\varepsilon'$-free for every $\varepsilon'\ge\varepsilon$ (this would solve the problem)? | |
| Apr 9, 2015 at 16:00 | comment | added | Ashutosh | Tangential remark: Note that you can always find a non null $Y \subseteq X$ avoiding mid points. This follows from: (R. Rado) One can partition $\mathbb{R}^n$ into countably many sets such that every piece is mid point free. See P. Komjath: Set theoretic constructions in Euclidean spaces, New Trends in Discrete and Computational Geometry (J. Pach, ed.), Springer, 1993, 303-325 | |
| Apr 9, 2015 at 15:50 | answer | added | Robert Israel | timeline score: 6 | |
| Apr 9, 2015 at 15:32 | comment | added | Mohammad Ghiasi | This is a beautiful question. | |
| Apr 9, 2015 at 14:58 | review | First posts | |||
| Apr 9, 2015 at 15:31 | |||||
| Apr 9, 2015 at 14:55 | history | asked | Mid | CC BY-SA 3.0 |