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Is there a characterization of measure zero subsets $A$ of $\mathbb R^n$, $n>1$ such that the set $A+A$ contains interior? Here $A+A$ is the set of points $\{ x+y \mid x, y\in A \}$.

Is it true that if the convex hull of the connected component of $A$ contains interior then so does $A+A$?

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Regarding Question #1.

There's one obvious dimensional obstruction (for example consider the Hausdorff dimension of A).

There is some research about it in the 1-dimensional case, such as the well-known theorem that K+K contains an interval, and related conjectures and works by Pallis, Furstenberg and Yoccoz. Even this one-dimensional theory is not complete as far as I know, so characterizing such a statement in larger dimensions would seem improbable now.

If say your set A is the product of two sets $A=A_{x} \times A_{y}$, such that $A_{x}+A_{x}$ contains an interval, and $A_{y}+A_{y}$ contains an interval, then one can have that the sumset of the product $A$ will contain a rectangle.

The one thing that can help you tackle the problem in larger dimensions, is the fact that you can sometimes say something smart about the projection of your set in a.e. direction (especially if your set is self-similar). For example this is the content of early works by Furstenberg (back in the 60s), and even some recent works (such as Hochman-Shmerkin - arXiv:0910.1956).

I hope that by taking two independent generic directions, you can say something smart about the projections (Hausdorff dimension? entropy estimates?), and then maybe one can use the one-dimensional theory to get that the sumset of the projection contains an interval. Of-course, it is not enough to get an interval in the projection, because your set might not be rectangular set itself, but maybe if your set if self-similiar, you can wiggle the pieces around to construct an inner rectangle.

By Masterand's theorem, I'm guessing that a resonable bet here would be $dim_{H}(A)>1$ (or maybe even take the upper packing dimension of $A$ to be bigger than $1$) for "nice enoguh" sets $A$, although I'm pretty sure this is open in general (because of Pallis' conjecture).

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  • $\begingroup$ What is "$K$" in this context? $\endgroup$
    – Will Sawin
    Commented May 21, 2012 at 18:08
  • $\begingroup$ K is the Cantor set (regular 1/3 Cantor set embedded in the unit interval). $\endgroup$
    – Asaf
    Commented May 21, 2012 at 21:19
  • $\begingroup$ I see that Stein and Shakarchi Vol 3 (Real Analysis) p. 48 Problem 7 is related to this question. $\endgroup$
    – spr
    Commented Jan 4, 2013 at 9:21
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For question #1, I don't know.

For question #2, the answer is no. Consider the edges of your favorite polyhedron in $\mathbb R^3$. They are connected. Their convex hull, the entire polyhedron, obviously has a nonempty interior. But $A+A$ is a finite union of two-dimensional parallelograms and thus cannot have interior in $\mathbb R^3$.

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  • $\begingroup$ @Will: I was surprised at your answer to question #2 until I realised that we read the question differently. I parsed OP's "so does $A+A$" to mean "the convex hull of the connected component of $A+A$ also contains interior" which would of course be trivially true. $\endgroup$
    – Willie Wong
    Commented May 21, 2012 at 9:17
  • $\begingroup$ The edges of polyhedron is 1 dimensional while the space is 3-dimensional. Can this difference of 2 dimensions be the reason? Precisely, is there a one-dimensional set $A$ for which $A+A$ can have interior in $\mathbb R^3$? Does the situation change if we substitute the edges of the polyhedron by something which does not contain an interval? $\endgroup$
    – spr
    Commented May 22, 2012 at 9:51
  • $\begingroup$ There is a natural map $A x A \to A + A$. If A is something such that "dimension" =is meaningful, such as a CW complex, then $A \times A$ has twice the dimension of $A$, and $A+A$ has no more than that dimension. Making $A$ curved will have no significant effect on the argument. Making $A$ totally disconnected will obviously make the connected components not have 3-dimensional convex hulls. I'm not sure which one you want. $\endgroup$
    – Will Sawin
    Commented May 22, 2012 at 16:03

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