Prove that for $S$ a maximal mean less subset, $|S|=\aleph$.

Question

We say that $S \subset \mathbb{R}$ is a mean less subset if for all $a \in S, b \in S$ we get that $(a+b)/2 \notin S$.

Prove that for $S$ a maximal mean less subset, $|S|=\aleph$.

Clearly we know that $|S|\leq\aleph$.

I need a direction how to prove that it can't be that $|S|< \aleph$.

Thank you for your help.

Answer 1

For $A\subseteq\Bbb R$ let $M(A)$ be the set of means of $S$:

$$M(A)=\left\{\frac12(a+b):a,b\in A\text{ and }a\ne b\right\}\,.$$

Clearly $|A|\le|M(A)|\le|A\times A|$, so $|M(A)|=|A|$ whenever $A$ is infinite.

Now suppose that $S\subseteq\Bbb R$, $S\cap M(S)=\varnothing$, and $\le|S|<|\Bbb R|$. The finite case is easy, so I will assume that $S$ is infinite. Let

$$D=\{|x-y|:x,y\in S\text{ and }x\ne y\}\,;$$

then $|D|=|S|$. (The argument is the same as for $|M(S)|$.) Let

$$T=\{x\pm d:x\in S\text{ and }d\in D\}\,;$$

$|T|=|S|$, so $|S\cup M(S)\cup T|=|S|<|\Bbb R|$. Let $x_0\in\Bbb R\setminus\big(S\cup M(S)\cup T\big)$, and let $S'=S\cup\{x_0\}$; clearly $S\subsetneqq S'$, and it’s not hard to see that.

$$M(S')=M(S)\cup\left\{\frac12(x+x_0):x\in S\right\}\,.$$

Suppose that $x\in S'\cap M(S')$; then there are distinct $y,z\in S'$ such that $x=\frac12(y+z)$.

• If $x\in S$, then $x\notin M(S)$, one of $y$ and $z$ is $x_0$, and the other is in $S$. Without loss of generality let $z=x_0$, and let $d=|x_0-x|$; $|x-y|=|x_0-x|$, so $d\in D$, and $x_0\in\{x-d,x+d\}\subseteq T$, which is impossible.
• If $x=x_0$, then $x_0\in M(S)$, which is also impossible.

Thus, $S'\cap M(S')=\varnothing$, $S'$ is a mean-less subset of $\Bbb R$, and $M$ is not a maximal mean-less subset of $\Bbb R$.