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May 16, 2023 at 18:19 vote accept Dominic van der Zypen
May 16, 2023 at 17:05 comment added Noah Schweber Or, just use the fact that there's a rational between $i(\beta)$ and $i(\beta+1)$ for each $\beta+1<\alpha$ to get a contradiction directly without porting over to $\mathbb{Q}$.
May 16, 2023 at 16:30 history answered Arno CC BY-SA 4.0