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Jun 27, 2020 at 14:41 comment added Andrés E. Caicedo @Andreas Yes, this is in essence the argument.
Jun 27, 2020 at 14:41 vote accept Zuhair Al-Johar
Jun 27, 2020 at 14:36 comment added Andreas Blass An alternative proof of the result about $F:\mathcal W(S)\to S$ without introducing $\sigma\mathcal M$: If there were such an $F$, then define, by recursion on ordinals $\alpha$, $G(\alpha)=F(\{G(\beta):\beta<\alpha\})$ and note that $G$ maps all the ordinals one-to-one into $S$. (To avoid mentioning the proper class of all ordinals, just restrict to $\alpha<$ Hartogs number of $S$.)
Jun 27, 2020 at 14:14 history answered Andrés E. Caicedo CC BY-SA 4.0