Timeline for Can There be a 1 dimensional Banach-Tarski paradox in the absence of choice
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2018 at 18:30 | history | edited | Josh F | CC BY-SA 3.0 |
changed notation for shifted subsets as per comment suggestion
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| Apr 19, 2018 at 18:19 | vote | accept | Josh F | ||
| Apr 19, 2018 at 18:17 | comment | added | Josh F | The argument I had in mind directly using the invariant mean characterization of amenability was decomposition $R$ into countable many copies of unit intervals. | |
| Apr 19, 2018 at 14:45 | answer | added | YCor | timeline score: 12 | |
| Apr 19, 2018 at 14:22 | answer | added | Will Sawin | timeline score: 16 | |
| Apr 19, 2018 at 14:19 | comment | added | YCor | I added the fa tag, because it's very likely to be provable with no choice, with a little functional analysis. | |
| Apr 19, 2018 at 14:18 | comment | added | YCor | Even with ZFC, the invariant mean characterization of $\mathbf{R}$ does not apply, since $[0,1]$ has mean zero for any invariant mean. One can use that $\mathbf{R}$ is supramenable, or use the trick of pushing forward such a paradoxical decomposition to a large enough circle. | |
| Apr 19, 2018 at 14:00 | history | edited | YCor |
edited tags
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| Apr 19, 2018 at 13:29 | comment | added | Nate Eldredge | Might I suggest writing $S_\alpha + r_\alpha$ instead of $r_\alpha(S_\alpha)$? | |
| Apr 19, 2018 at 12:04 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
removing superfluous word
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| Apr 19, 2018 at 11:24 | history | edited | Josh F | CC BY-SA 3.0 |
clarified question
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| Apr 19, 2018 at 2:41 | review | First posts | |||
| Apr 19, 2018 at 3:00 | |||||
| Apr 19, 2018 at 2:36 | history | asked | Josh F | CC BY-SA 3.0 |