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Mar 18, 2015 at 18:35 history edited Dominic van der Zypen CC BY-SA 3.0
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Mar 18, 2015 at 15:05 comment added Ramiro de la Vega Noah is right. You should erase the ",$n\geq2$" to make it work.
Mar 18, 2015 at 6:59 comment added Noah Schweber I don't think this works - as I wrote in my comment, I don't think there's an injection $f$ from $\bigcup\mathcal{A}$ to $\mathcal{A}$ satisfying $a\in f(a)$, since 0 and 1 must both be sent to the same place.
Mar 18, 2015 at 6:50 vote accept CommunityBot
Sep 2, 2015 at 16:36
Mar 18, 2015 at 6:42 comment added Noah Schweber If you mean $\{\{0, 1\}\}\cup . . .$, I don't think this works - what is the $f$? Otherwise - if we treat numbers as ordinals - 0 can't be in $\mathcal{A}$ since $\mathcal{A}$ should consist of nonempty sets only.
Mar 18, 2015 at 6:41 history answered Dominic van der Zypen CC BY-SA 3.0