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Jul 18, 2022 at 23:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jun 18, 2022 at 22:49 answer added Noah Schweber timeline score: 1
Jun 18, 2022 at 22:22 comment added Andreas Blass Not a genuine example but a conditional one: If $V=L$ then there is a $\Delta^1_2$ well-ordering of the reals with order-type $\omega_1$. Such a well-ordering, as a subset of $\mathbb R^2$, is not Lebesgue measurable and therefore not in the $\sigma$-algebra generated by analytic sets.
Jun 18, 2022 at 21:49 history edited YCor CC BY-SA 4.0
removed capitals from title, added tag
Jun 18, 2022 at 21:42 comment added John Levy @Wojowu Sorry, corrected!
Jun 18, 2022 at 21:41 history edited John Levy CC BY-SA 4.0
added 4 characters in body
Jun 18, 2022 at 20:33 comment added Wojowu Are you sure you wrote the question correctly? There is some awkward double negation in there, and that question seems different from what is mentioned in the title of the question.
Jun 18, 2022 at 20:15 history asked John Levy CC BY-SA 4.0