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Feb 8, 2021 at 19:59 history edited Vincent Granville CC BY-SA 4.0
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Feb 5, 2021 at 17:12 history edited Vincent Granville CC BY-SA 4.0
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Feb 5, 2021 at 16:45 comment added Vincent Granville I'll try to fix my wording, but true, I am not a native speaker.
Feb 5, 2021 at 16:43 comment added Asaf Karagila Yes, but you're not reading my [previous] comment.
Feb 5, 2021 at 16:43 comment added Vincent Granville @Asaf: unless I am mistaken, a set can have measure 0 yet be non countable. For instance, consider the set of all real numbers and their representation in base 2. For each real, add a 0 in the binary digits in positions 1, 3, 5, and so on. The transformed numbers are in bijection with real numbers, but none of them is a normal number. So it is an (uncountable) subset of non-normal numbers. And non-normal numbers have measure 0, so that set also has measure 0.
Feb 5, 2021 at 11:13 comment added Asaf Karagila Since every countable set has measure 0, saying "either X is countable, or X has measure 0" is kind of redundant (and either-or reads to me, as a non-native speaker anyway, as a dichotomy, so exactly one of the options is true).
Feb 5, 2021 at 4:10 history became hot network question
Feb 5, 2021 at 0:08 vote accept Vincent Granville
Feb 4, 2021 at 20:53 answer added Wojowu timeline score: 10
Feb 4, 2021 at 20:11 comment added Vincent Granville It would be interesting to see which real numbers $\alpha$ have $R(\alpha)=R(\pi)$, assuming $R(\pi)$ exists.
Feb 4, 2021 at 20:06 history asked Vincent Granville CC BY-SA 4.0