Timeline for Partition of R into midpoint convex sets
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2010 at 11:52 | comment | added | domotorp | I want my name with all capitals and the comment attached to the main question, so that it becomes well visible from far | |
| Apr 24, 2010 at 8:17 | vote | accept | filipm | ||
| Apr 24, 2010 at 8:17 | vote | accept | filipm | ||
| Apr 24, 2010 at 8:17 | |||||
| Apr 24, 2010 at 8:17 | comment | added | filipm | domotorp: "lemma. if 0 is red and 1 is blue, then all the pos integers are blue. now take any three numbers such that x<y<z and x and z are red, while y is blue. if such a triple does not exist, we have a trivial partition. linearly transform the rationals such that x=0 and y and z are integers, k and k+n. then the multiples of k are all blue, numbers of the form k+in are all red (using the lemma again), but this is a contradiction at k+kn." | |
| Apr 24, 2010 at 8:16 | comment | added | filipm | Thanks for the answer. So, we can use a trivial partition of $\mathbb Q$ to generate non-trivial partitions of $\mathbb R$. However, it's easy to see that $\mathbb Q$ itself doesn't admit non-trivial partitions, e.g. in the next comment I give an approach by mr domotorp. | |
| Apr 23, 2010 at 12:47 | comment | added | Wadim Zudilin | @Joel: Yes, the trick is well generalisable. | |
| Apr 23, 2010 at 12:46 | comment | added | Wadim Zudilin | Thanks, Harald, for this correction. I am mostly unhappy with the above solution because, in a certain sense, it is "almost trivial": we replace the usual $\mathbb R$-order by the one induced by a $\mathbb Q$-basis... | |
| Apr 23, 2010 at 12:42 | comment | added | Joel David Hamkins | It seems you can replace Q in this argument with larger subfields of R and gain even greater degrees of convexity in a nontrivial partition. | |
| Apr 23, 2010 at 12:39 | comment | added | Harald Hanche-Olsen | @Wadim: Obviously not, since not assuming AC does not guarantee that AC is not true. You should really ask if “all partitions are trivial” is consistent with ZF without AC. (I don't know the answer.) | |
| Apr 23, 2010 at 12:31 | comment | added | Wadim Zudilin | I am not an AC fan. Is it possible to show that without AC a desired partition is necessarily trivial? | |
| Apr 23, 2010 at 12:28 | comment | added | Steven Gubkin | Nice solution. I had just come up with the same right before you posted. Now my question is: Is AC needed for a solution? | |
| Apr 23, 2010 at 12:18 | history | answered | Thomas Kragh | CC BY-SA 2.5 |