# Steinhaus theorem and Hausdorff dimension

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.

• There's extensive literature on the case of Cantor sets, based on Furstenberg's conjecture (now Hochman's theorem).
– Asaf
Sep 1, 2023 at 3:09
• @Asaf Could you be more specific? Saying "there is extensive literature" without saying what literature you have in mind and without saying what you mean by Furstenberg conjecture is not very useful. You could place it as an answer. Sep 1, 2023 at 4:00
• Furstenberg's conjecture is about dimension of intersections of (self-similar sets). There are several papers by Hochman around that conjecture, in particular in annals.math.princeton.edu/wp-content/uploads/… - Hochman and Shmerkin show dimension growth for sumsets of "disjoint" Cantor sets.
– Asaf
Sep 1, 2023 at 4:20

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.

• Wow! Very beautiful and unexpected result. Thank you. Sep 1, 2023 at 4:01

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.

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.

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