Timeline for Does ZF bound countable unions of countable sets?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2012 at 12:13 | vote | accept | Colin McLarty | ||
| Oct 15, 2012 at 8:02 | history | edited | Asaf Karagila♦ |
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| Oct 15, 2012 at 8:00 | answer | added | Asaf Karagila♦ | timeline score: 8 | |
| Oct 15, 2012 at 7:47 | comment | added | Asaf Karagila♦ | Michael, no. It should be $\aleph_1$. If $\aleph_1$ is singular then it is the countable union of countable ordinals; but $\aleph_2$ (even if singular) can never be the countable union of countable ordinals. | |
| Oct 15, 2012 at 7:21 | comment | added | Michael Greinecker | I think that should be $\aleph_0$ in the first sentence. | |
| Oct 15, 2012 at 4:45 | history | asked | Colin McLarty | CC BY-SA 3.0 |