Steinhaus theorem and Hausdorff dimension

Question

Assume for simplicity that sets $A_i\subset\mathbb{R}$ are compact. If $A_1$ and $A_2$ have positive length, then $A_1+A_2$ contains an interval. That is a variant of the classical Steinhaus theorem and it easily follows by looking at neighborhoods of a density point of $A_1$ and a density point of $A_2$.

Question. Let $\alpha\in (0,1)$. Is it true that then there is $k$ such that if $\mathcal{H}^\alpha(A_i)>0$, $i=1,2,\ldots,k$, then $A_1+A_2+\ldots+A_k$ contains an interval?

Here $\mathcal{H}^{\alpha}$ stands for the Hausdorff measure.

I am also interested in a version of this question where $A_1=A_2=\ldots=A_k$ i.e., if we add a set $A$ with $\mathcal{H}^\alpha(A)>0$ to itself $k$ times: $A+A+\ldots+A$.

If this is not true, but simiar results are true, I would be interested in references. I am particulary interested in sets in a line.

Answer 1

The answer to the question is negative. Körner in Hausdorff dimension of sums of sets with themselves and Schmeling-Shmerkin in On the dimension of iterated sumsets showed that for any increasing sequence $\alpha_1 < \alpha_2 < \ldots < 1$, there is a compact set $A$, s.t. for every $k$, we have $\dim_{H} k A = \alpha_k$, where $kA = A + A + \ldots + A$ --- Minkowski sum of the set with itself $k$ times, and $\dim_H$ is the Hausdorff dimension of the set. In particular, choosing any sequence such that $\dim_H kA < 1$ for all $k$, none of the sum sets $kA$ could contain an interval.

See also the recent paper Dimension growth for iterated sumsets for some conditions under which we have dimension growth $\dim_H (A + A) > \dim_H(A)$. Unfortunately, it seems that the bounds there still fall short from implying that for some $k$, $\dim_H k A = 1$, much less providing quantitative bound on $k$.

On the positive side, Marstand in "Some Fundamental Geometrical Properties of Plane Sets of Fractional Dimensions" showed that for any given pair of sets $A,B$, and almost every angle $\theta$, if we write $\lambda_1 = \cos \theta, \lambda_2 = \sin \theta$, then $$ \dim_H (A_{\lambda_1} + B_{\lambda_2}) = \min(\dim_H(A \times B),1) \geq \min(\dim_H(A) + \dim_H(B), 1),$$ where $A_{\lambda}$ is the dilatation $A_{\lambda} := \{ \lambda x : x \in A\}$. In fact a slightly stronger statement holds: if $\dim_H (A) = \alpha$, there is a sequence of parameters $\lambda_1, \lambda_2, \ldots, \lambda_k > 0$ for $k = \lceil 1/\alpha \rceil + 1$, such that for $B := A_{\lambda_1} + \ldots + A_{\lambda_k}$ not only $\dim_H(B)=1$, but indeed the 1-dimensional measure of this set is positive, and therefore $B+B$ contains an interval (a generic sequence of $\lambda_i$ will do the job). Unfortunately, this result does not say anything about any specific sequence of $\lambda_i$ (of which $\lambda_1 = \ldots = \lambda_k$ is the most interesting). A more modern exposition of this theorem can be found in Chapter 9 of "Geometry of sets and measures in Euclidean spaces" by Pertti Mattila.

Answer 2

Here is a quite short example to show that you question cannot have a positive answer.

Assume that $V$ is a Sacks extension of constructible universe $L$. Then the set of constructible reals $A=(\mathbb{R})^L$ is nonmeasurable and so has Hausdorff dimension 1. But for any $k$, $\sum_k A\subseteq A$ has inner measure 0 and so has no subinterval.

But to construct a counterexample under $ZFC$, we need more knowledge from logic. The rough idea is to construct a nonmeasurable ideal of hyperarithmetic degrees.

Answer 3

Lemma 2.7 of [1] says that if $A$ is a compact subset of a separable compact abelian group $G$ with Haar measure 0, then there is a compact set $B\subset G$ with positive Haar measure such that $A+B$ is nowhere dense. The same proof works for $\mathbb R$, so for any compact $A\subset \mathbb R$ of Hausdorff dimension $1$ and Lebesgue measure $0$, there is a compact set $B\subset \mathbb R$ of positive Lebesgue measure such that $A + B$ does not contain an interval.

[1] Griesmer, John T., Sumsets of dense sets and sparse sets, Isr. J. Math. 190, 229-252 (2012). ZBL1312.11007.

Answer 4

In the paper by Feng and Wu, they introduced a notion of thickness, that generalized the Newhouse thickness in the line. They showed that if a set has positive Feng-Wu thickness, then the sumset after k times will have an interior (k is also estimated in the paper). They also showed that all self-similar sets has positive Feng-Wu thickness. This partially answered your second question that something similar is true.

DE-JUN FENG AND YU-FENG WU, "On the Arithmetic Sums of Fractal sets in ${\mathbb R}^d$, J. Lond. Math. Soc. 104(2021), no. 1, 35-65.

https://arxiv.org/pdf/2006.12058.pdf