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?
Best Answer
We can find Cantor-like sets of Hausdorff dimension arbitrarily close to $1$ which satisfy your property.
Lemma: If a subset $S \subseteq \{1, 2, \dots, n\}$ of cardinality $|S| = m$ has no length-$3$ arithmetic progressions, then we can find a subset $K \subsetneq [0, 1]$ with a Hausdorff dimension of:
$$ \dfrac{\log{m}}{\log(2n-1)} $$
In particular, the Cantor ternary set is the special case of applying this construction to $S = \{1, 2\}$.
Proof: We apply a Cantor-like construction, replacing the interval $K_0 = [0, 1]$ with $m$ intervals:
$$ K_1 = \bigcup \{ [\dfrac{2s-2}{2n-1}, \dfrac{2s-1}{2n-1}] : s \in S \} $$
and iterating in the obvious way to produce a sequence $K_0 \supsetneq K_1 \supsetneq K_2 \supsetneq \dots$. Define $K$ to be the limit (intersection) of these sets.
Now given any two points $x, y \in K$, we consider the first iteration $K_n$ of the process which puts $x,y$ in different intervals. Then I claim that $\frac{x+y}{2}$ is not in $K_n$.
Due to the recursive structure of $K$, we can wlog assume $n = 1$. Hence we just need to show that if $x$ and $y$ belong to different intervals in $K_1$, their mean does not belong to $K_1$. This is where we make use of the $3AP$-less-ness of $S$.
For simplicity of notation, we'll scale $K_1$ up by $2n - 1$ and suppose that:
$$ x \in [2s - 2, 2s - 1], y \in [2t - 2, 2t - 1] $$
where $s,t \in S$ and wlog $s < t$.
Then the mean of $x$ and $y$ must consequently satisfy:
$$ \dfrac{x + y}{2} \in [s + t - 2, s + t - 1] $$
If $s + t$ is odd, this interval doesn't belong to $K_1$ by parity. If $s + t$ is even, then this interval doesn't belong to $K_1$ because $\frac{1}{2}(s + t) \notin S$ by the $3AP$-less-ness of $S$.
The result follows.
Theorem: For every $\varepsilon > 0$, we can find a set $K$ of Hausdorff dimension greater than $1 - \varepsilon$
Proof: By Lemma 1, we just need to find a $m$-subset of $[n]$ with no three-term arithmetic progression and which satisfies:
$$ \dfrac{\log{m}}{\log(2n-1)} > 1 - \varepsilon$$
or equivalently:
$$ \dfrac{\log(\frac{m}{2n - 1})}{\log(2n - 1)} > -\varepsilon$$
or equivalently:
$$ \dfrac{m}{2n - 1} > (2n - 1)^{-\varepsilon} $$
It is thus sufficient to find a $3AP$-less set with density $\delta := \dfrac{m}{n}$ satisfying:
$$ \delta \geq 2(2n - 1)^{-\varepsilon} $$
Such large $3AP$-less sets exist. See Theorem 3.1 (Behrend, 1946) of the following paper:
http://wiki.math.toronto.edu/TorontoMathWiki/images/2/2d/Expo_paper.pdf
QED #ReductionToAKnownProblem