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Jan 27, 2018 at 10:46 comment added Asaf Karagila Indeed. Of course, if you want $\aleph_2$ to also be regular, an inaccessible is needed.
Jan 27, 2018 at 7:15 comment added Trevor Wilson Ah, right. I should have seen the analogy with getting $\aleph_1 \not\le \mathbb{R}$.
Jan 27, 2018 at 7:14 vote accept Trevor Wilson
Jan 27, 2018 at 7:00 history answered Asaf Karagila CC BY-SA 3.0