TITLE: Hausdorff dimension of a Cantor-like set
QUESTION [6 upvotes]: Suppose $K$ is a subset of $[0,1]$ with the following property: for almost $x,y \in K$, we have
$$\frac{x+y}{2} \not\in K.$$
(Here, "almost in $K$" means "in $K$ except for a countable subset").
Such a set must have holes, and the Cantor tridiagonal set has this property. What can we say about the Hausdorff dimension of $K$? Is it true that it is less than $\ln 2 / \ln 3$?
Can we even say that the Hausdorff measure $H^{\ln 2/ \ln 3}(K)$ is finite?

REPLY [4 votes]: In Construction of 1-dimensional subsets of the reals not containing similar copies of given patterns Tamás Keleti constructs a compact set of Hausdorff dimension $1$ which contains no similar copy of any out a given countable number of $3$-element sets. In particular, there is a compact set of Hausdorff dimension $1$ which does not contain any arithmetic progressions.
His construction does not use Behrend's example at all, and it is intrinsic to the reals as it depends on a multi-scale argument, so it is substantially different from Adam Goucher's nice solution.
On the other hand, an easy application of the Lebesgue density theorem shows that any set of positive Lebesgue measure contains an arithmetic progression of length $3$ (and of any finite length).