Let $\ A\subset R $ have the following characteristic:
For all $\ a,b \in A$ , $\ \frac{a+b}{2} \notin A$.
Prove that there exists a maximal set A. Prove its cardinality is $\ \aleph $.
The first part is relatively simple using Zorn's Lemma, taking any chain with the inclusion relation, of sets with the said characteristic, and binding them above by their union.
As for the second part. Let $\ max{A}=M $. $\ M\subset R $ so $\ |M| \leq \aleph $ .
My question is, how can I find a set of Cardinality $\ \aleph $ with the stated characteristic (or prove one exists) to be a lower bound for M?
Clearly $M$ is infinite. So $|M\times M|=|M|$. By the maximality of $M$, for all $x\in \mathbb{R}\backslash M$, there is an element $m_x\in M$ such that $\frac{x+m_x}{2}\in M$. For each $x\in \mathbb{R}\backslash M$, fix a choise of $m_x$. We can then define a function $f: \mathbb{R}\backslash M \to M\times M$ so that $f(x)=(m_x, \frac{x+m_x}{2})$. If $x,y \in \mathbb{R}\backslash M$ and $f(x)=f(y)$, then $m_x=m_y$ and $\frac{x+m_x}{2}=\frac{y+m_y}{2}$, so $x=y$. This means that $f$ is injective. Therefore $$ |\mathbb{R}\backslash M| \leq |M\times M| = |M|.$$ So we have $|M|=|\mathbb{R}\backslash M| + |M| = |\mathbb{R}|$.