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?

  • $\begingroup$ By $\aleph$, are you referring to $\aleph_0$, the least infinite cardinal? $M$ being a subset of $\Bbb R$ is not sufficient to prove that $|M| \le \aleph_0$. It is sufficient to prove that $|M| \le |\Bbb R|$, but using $\aleph$ to represent this cardinality runs counter to the established convention for naming cardinals. $\Bbb R$ is often called "$c$", since its place in the $\aleph$ naming structure is indeterminant. $\endgroup$ – Paul Sinclair Jun 22 '16 at 0:10
  • $\begingroup$ By $\ \aleph$ I mean $\ 2^{\aleph_0} = | \Bbb R | = c$. $\endgroup$ – Jonathan Jun 22 '16 at 6:20

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 $|M|=|\mathbb{R}\backslash M| + |M| = |\mathbb{R}|$.

  • $\begingroup$ I understand your argument, but it seems to me it holds only if $\ |\mathbb{R}\backslash M| = \aleph $. How do I know it is not finite? $\endgroup$ – Jonathan Jun 22 '16 at 15:23
  • $\begingroup$ @Jonathan - Since $|\mathbb{R}\backslash M| + |M| = |\mathbb{R}|$, if $|\mathbb{R}\backslash M| < |\mathbb{R}|$, then $|M| = |\mathbb{R}|$. And since $|\mathbb{R}\backslash M| \le |M|$, if $|\mathbb{R}\backslash M| = |\mathbb{R}|$, then so does $|M|$. $\endgroup$ – Paul Sinclair Jun 22 '16 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.