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Sep 8, 2014 at 22:19 vote accept François Le Maître
Sep 8, 2014 at 21:56 answer added François Le Maître timeline score: 1
Jun 11, 2014 at 7:23 comment added François Le Maître Indeed the equality up to measure zero smooths out the complexity, I added a proof of the fact that this set is analytic.
Jun 11, 2014 at 7:16 history edited François Le Maître CC BY-SA 3.0
added proof that this set is analytic
Jun 11, 2014 at 0:11 comment added Joel David Hamkins Could you explain your remark about it being analytic? For example, the set of Borel codes for Borel functions $f:\mathbb{R}\to\mathbb{R}$ with countable range seems to me to be $\Sigma^1_2$. Are you claiming it is actually $\Sigma^1_1$? Or is it the equivalence that smooths out this complexity?
Jun 10, 2014 at 13:36 review First posts
Jun 10, 2014 at 13:37
Jun 10, 2014 at 13:11 history asked François Le Maître CC BY-SA 3.0