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Apr 12 at 16:36 comment added Zuhair Al-Johar Thank you very much.
Apr 12 at 16:35 comment added Asaf Karagila Oh, I see what you're asking for now. Sure, that can also work. But I was aiming for correct, not optimal.
Apr 12 at 16:33 comment added Zuhair Al-Johar As far as I understand what we need is for $A$ to be strictly larger than every element of $V_{\omega+\omega}$, and so we can take $A$ to be of size $V_{\omega+\omega}$.
Apr 12 at 16:31 comment added Asaf Karagila If we started with $A$ strictly larger, how would that come to be?
Apr 12 at 16:30 comment added Zuhair Al-Johar I meant of size $V_{\omega+\omega}$, sorry for the typo.
Apr 12 at 15:54 comment added Asaf Karagila It's not well founded... I don't understand your question.
Apr 11 at 17:01 comment added Zuhair Al-Johar Why we don't take the size of $A$ to be $V_{\omega+\omega}$?
Sep 16, 2020 at 20:14 vote accept Zuhair Al-Johar
Sep 16, 2020 at 20:04 history answered Asaf Karagila CC BY-SA 4.0