Borel set plus a closed set = Borel - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:11:34Z http://mathoverflow.net/feeds/question/48571 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48571/borel-set-plus-a-closed-set-borel Borel set plus a closed set = Borel Wishiwere Smarter 2010-12-07T16:03:05Z 2010-12-07T17:37:00Z <p>Hi,</p> <p>Let $R$ be equipped with the usual Borel structure. Let $F$ be a Borel subset and $E$ be a closed subset of $R$. Then $F+E=(f+e: f\in F, e \in E )$ is Borel? If yes, is it true for any locally compact topological group? Thanks in advance.</p> http://mathoverflow.net/questions/48571/borel-set-plus-a-closed-set-borel/48577#48577 Answer by gowers for Borel set plus a closed set = Borel gowers 2010-12-07T16:25:24Z 2010-12-07T16:55:18Z <p>At least in $R^2$ it's false, and probably in R too.</p> <p>There exist closed subsets of $R^2$ that project to non-Borel. So if you take such a set and add it to the y-axis then you'll get a non-Borel set too.</p> <p>There's probably some general nonsense that would allow you to transfer this result to R, but I don't know it.</p> <p>Edit: Actually, perhaps I do. The above argument works in $\mathbb{N}^{\mathbb{N}}$, which is isomorphic to its square and can be embedded additively into $\mathbb{R}$. I haven't checked that this works, but it feels as though it should.</p> http://mathoverflow.net/questions/48571/borel-set-plus-a-closed-set-borel/48579#48579 Answer by Andrey Rekalo for Borel set plus a closed set = Borel Andrey Rekalo 2010-12-07T17:23:05Z 2010-12-07T17:37:00Z <p>No. Erdös and Stone showed that the sum of two subsets $E$, $F\subset\mathbb R$ may not be Borel even if one of them is compact and the other is $G_\delta$ (see <a href="http://www.jstor.org/pss/2037209" rel="nofollow"><em>"On the Sum of Two Borel Sets"</em></a>, Proc. Am. Math. Soc., Vol. 25, (1970), pp. 304-306). </p> <p>Their argument works for every connected locally compact (or abelian) topological group with a complete metric.</p>