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Hi,

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.

flag	asked Dec 7 2010 at 16:03 Wishiwere Smarter 163●3

Andrey Rekalo · Accepted Answer · 2010-12-07 17:37:00Z UTC

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 "On the Sum of Two Borel Sets", Proc. Am. Math. Soc., Vol. 25, (1970), pp. 304-306).

Their argument works for every connected locally compact (or abelian) topological group with a complete metric.

gowers · Answer 2 · 2010-12-07 16:55:18Z UTC

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At least in $R^2$ it's false, and probably in R too.

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.

There's probably some general nonsense that would allow you to transfer this result to R, but I don't know it.

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.

edited Dec 7 2010 at 16:55

answered Dec 7 2010 at 16:25

gowers
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@gowers : Actually, I do not think this is correct. (It is if instead of the reals we work with the irrationals.) To get non-Borel sets it suffices to project the complement (in ${\mathbb R}^2$) of the projection of a closed subset of ${\mathbb R}^3$. – Andres Caicedo Dec 7 2010 at 16:56

Are you saying that a projection of a closed set in $\mathbb{R}^2$ has to be Borel? – gowers Dec 7 2010 at 17:43

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@gowers : Yes. The general result in ${\mathbb N}^{\mathbb N}$ that projections of closed sets are not Borel does not transfer intact to ${\mathbb R}$. – Andres Caicedo Dec 7 2010 at 17:53

Ah yes, a closed subset of $\mathbb{R}^2$ can be written as a union of countably many compact sets, so it obviously can't work. – gowers Dec 7 2010 at 18:16

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It is a source of confusion, since in descriptive set theory most results are independent of the specific Polish space we use. I guess I could add it to the list of common mistakes... – Andres Caicedo Dec 7 2010 at 18:53

Borel set plus a closed set = Borel

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