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Jan 4, 2016 at 22:15 answer added William timeline score: 7
Jan 4, 2016 at 18:34 comment added Ohad Drucker Thanks! So should I understand from your comments that if CH + "no L - Cohen reals", then after adding $\aleph_2$ random reals, $\mathbb{V}[G] \models \neg CH$ + "no L - Cohen reals" ?
Jan 4, 2016 at 16:28 comment added Asaf Karagila One key point here is that if there are no Cohen reals over L, then $\omega_1$ is computed correctly in $L$.
Jan 4, 2016 at 15:35 comment added Ohad Drucker Does adding \aleph_2 random reals work? Or \aleph_2 Sacks reals? And if so, how do one show that?
Jan 4, 2016 at 15:15 comment added Thomas Benjamin Use random reals.
Jan 4, 2016 at 14:57 history asked Ohad Drucker CC BY-SA 3.0