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Jul 8, 2018 at 13:27 vote accept Stefan Mesken
Jul 5, 2018 at 1:52 history edited Stefan Mesken CC BY-SA 4.0
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Jul 4, 2018 at 20:23 comment added Asaf Karagila @Miha: Of you're absolutely right. I guess that what happens when it's 21:00 and you still haven't had breakfast... :)
Jul 4, 2018 at 20:12 comment added Miha Habič @AsafKaragila If $0^\sharp$ exists then it is in HOD, so I would assume that $\mathrm{HOD}^{L[0^\sharp]}=L[0^\sharp]$.
Jul 4, 2018 at 19:14 comment added Asaf Karagila What is $\mathrm{HOD}^{L[0^\#]}$, by the way?
Jul 4, 2018 at 18:54 history edited Stefan Mesken CC BY-SA 4.0
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Jul 4, 2018 at 18:49 history answered Stefan Mesken CC BY-SA 4.0