Timeline for Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2022 at 22:52 | comment | added | Elliot Glazer | That still doesn't work: for two disjoint pieces $A$ and $B,$ the three sets $A,$ $B,$ and $A \cup B$ add to 0 in the vector space. | |
| Mar 29, 2022 at 22:45 | comment | added | Andreas Blass | @ElliotGlazer Thanks for the correction. Instead of "symmetric difference of these pieces", I should have said "symmetric difference of these unions". | |
| Mar 29, 2022 at 21:49 | comment | added | Elliot Glazer | "In particular, no finite, nonempty, symmetric difference of these pieces is measurable." The pieces themselves are only a set of cardinality $\mathfrak{c}$ and there are many linear dependencies among the $2^{\mathfrak{c}}$ unions of them. That being said, you're right the dimension is $2^{\mathfrak{c}},$ since cardinality equals dimension in infinite vector spaces over a finite field. | |
| Jul 16, 2012 at 23:47 | comment | added | Gene S. Kopp | Nice argument. Thanks for the answer! | |
| Jul 16, 2012 at 23:33 | vote | accept | Gene S. Kopp | ||
| Jul 16, 2012 at 21:02 | history | answered | Andreas Blass | CC BY-SA 3.0 |