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Jul 20, 2022 at 0:12 history edited Sam Roberts CC BY-SA 4.0
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Aug 12, 2020 at 8:00 vote accept Sam Roberts
Aug 7, 2020 at 8:16 answer added Mohammad Golshani timeline score: 14
Aug 5, 2020 at 19:51 comment added Sam Roberts oh, haha! (I use weaklymeasurable as a handle on instagram.) (1) is fixed. So, I think Foreman-Woodin was just $2^\lambda> \lambda^+$ and then Woodin got $2^\lambda = \lambda^{++}$.
Aug 5, 2020 at 19:49 history edited Sam Roberts CC BY-SA 4.0
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Aug 5, 2020 at 19:45 comment added Monroe Eskew Two questions: (1) What is weakly measurable? (2) Remind me what does the original Foreman-Woodin model satisfy?
Aug 5, 2020 at 17:43 history edited Sam Roberts CC BY-SA 4.0
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Aug 5, 2020 at 14:55 history edited Martin Sleziak
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Aug 5, 2020 at 14:29 history asked Sam Roberts CC BY-SA 4.0