Timeline for Non-completeness of the Borel-Lebesgue measure and countable choice
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2016 at 0:11 | comment | added | Joel David Hamkins | The Cantor set is equinumerous with the reals, and so they have the same size power set. Since we can map the reals surjectively onto the Borel subsets of the Cantor set, but not onto the full power set, there must be ($2^{\mathfrak c}$) many subsets of the Cantor set that are not Borel. | |
| Mar 25, 2016 at 23:38 | vote | accept | Michael | ||
| Mar 25, 2016 at 23:38 | comment | added | Michael | Great answer yet again. Just a supplementary question : how can this implies that there is a non-Borel set included in the Cantor set ? | |
| Mar 25, 2016 at 9:17 | comment | added | Asaf Karagila♦ | To complete the argument about the non-completeness of the measure, this implies there is a subset of the Cantor set which is not Borel, and therefore the measure is not complete. | |
| Mar 25, 2016 at 5:43 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |