Timeline for The paradox with the first uncountable ordinal
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2021 at 16:00 | review | Close votes | |||
| Aug 25, 2021 at 17:34 | |||||
| Apr 7, 2016 at 1:15 | history | edited | David Feldman | CC BY-SA 3.0 |
Fixed wording
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| Jun 7, 2013 at 0:05 | answer | added | Noah Schweber | timeline score: 4 | |
| Jun 6, 2013 at 15:59 | comment | added | The User | @Ilya Because every bounded subset of a set of order type $\omega_1$ is countable. | |
| Jun 6, 2013 at 14:50 | vote | accept | Dan | ||
| Jun 6, 2013 at 14:48 | comment | added | SBF | Can you explain, please, why the set $M(a)$ is countable? | |
| Jun 6, 2013 at 14:22 | comment | added | The User | If CH fails, it does not work: Every Lebesgue-measurable subset with cardinality less than the continuum is a null set: mathoverflow.net/questions/8972/… | |
| Jun 6, 2013 at 14:12 | history | edited | The User |
edited tags
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| Jun 6, 2013 at 14:05 | answer | added | The User | timeline score: 1 | |
| Jun 6, 2013 at 14:04 | answer | added | Andreas Blass | timeline score: 8 | |
| Jun 6, 2013 at 14:01 | comment | added | François G. Dorais | Ah, I see, I had missed that bit and only read $M \subset R$. | |
| Jun 6, 2013 at 13:59 | comment | added | The User | @François Yes. I do not see any possible different interpretation of his words. | |
| Jun 6, 2013 at 13:57 | comment | added | Andreas Blass | Dan assumed CH when he wrote the first sentence of the question, where he supposed that the interval $(0,1)$ can be well-ordered as the first uncountable ordinal. | |
| Jun 6, 2013 at 13:52 | comment | added | François G. Dorais | @User: So Dan is assuming CH? | |
| Jun 6, 2013 at 13:47 | comment | added | The User | @François Hm? The standard probability measure (Lebesgue measure) on the unit interval is not null. | |
| Jun 6, 2013 at 13:44 | answer | added | Denis | timeline score: 11 | |
| Jun 6, 2013 at 13:42 | comment | added | François G. Dorais | @User: So $M$ is Lebesgue measurable and not null? | |
| Jun 6, 2013 at 13:41 | comment | added | The User | @François I guess he uses Lebesgue measure on $M$. | |
| Jun 6, 2013 at 13:36 | comment | added | François G. Dorais | To pick elements from $M$ at random you need to have a probability measure on $M$ to sample from. What is it? | |
| Jun 6, 2013 at 13:31 | history | asked | Dan | CC BY-SA 3.0 |