Timeline for A set of positive measure with cardinality less than that of the continuum?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2013 at 12:45 | comment | added | Alexander Pruss | Another proof is here: mathoverflow.net/questions/125563/… | |
| Mar 25, 2013 at 17:28 | comment | added | Sean Eberhard | @AndreasBlass I should've added that by inner regularity you can find a closed subset also of positive measure. Now apply that argument. | |
| Mar 25, 2013 at 17:24 | comment | added | Andreas Blass | @Sean: Concerning your comment about repeated subdivision, did you intend the "continuum-many points" in the last sentence to be the limits of the subdivision points? If so, how do you know they're in $X$. If not, which continuum-many points did you mean? | |
| Mar 25, 2013 at 14:51 | comment | added | Sean Eberhard | Aha, yes, that question includes information about the non-measurable case, which I ignored. | |
| Mar 25, 2013 at 14:49 | comment | added | Sean Eberhard | Alternatively, if $X\subset[0,1]$ has positive measure, find $x\in[0,1]$ such that $[0,x]\cap X$ and $[x,1]\cap X$ both have positive measure. Then find $x_0$ and $x_1$ such that $0<x_0<x<x_1<1$ and $X$ has positive measure in each of the intervals $[0,x_0]$, $[x_0,x]$, $[x,x_1]$, $[x_1,1]$. Continue subdividing, and find continuum-many points in $X$. | |
| Mar 25, 2013 at 14:48 | comment | added | Alexander Pruss | Link was bad: mathoverflow.net/questions/8972/… | |
| Mar 25, 2013 at 14:48 | comment | added | Alexander Pruss | Thanks! I was about to delete my question in light of <a href="mathoverflow.net/questions/8972/…> (which has good information on related stuff) and then I saw your very quick answer. | |
| Mar 25, 2013 at 14:46 | vote | accept | Alexander Pruss | ||
| Mar 25, 2013 at 14:44 | history | answered | Sean Eberhard | CC BY-SA 3.0 |