## May 1990 #3

Problem Statement: For $A\subset \mathbb{R}$, let $A'$ denote the set of cluster points of $A$. Show there exists no set $A\subset \mathbb{R}$ such that $A'=\mathbb{Q}$.

Proof: Suppose that there is a set $A\subset \mathbb{R}$ such that $A'=\mathbb{Q}$. Let $r\in \mathbb{R}$ such that $r$ is irrational.

Let $\delta>0$ and consider the open neighborhood $(r-\delta,r+\delta)\subset\mathbb{R}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ it follows that there is some rational $q\in (r-\delta,r+\delta)$. Since our assumption was that $A'=\mathbb{Q}$ it follows that since $(r-\delta,r+\delta)$ is a neighborhood of $q$ there is an infinite subset of $A$ contained in that neighborhood. So, for every $\delta>0$ it follows that $(r-\delta,r+\delta)$ contains an infinite subset of $A$ and so $r\in A'$. But this is a contradiction since $r$ is irrational. Thus our original assumption must have been wrong.

Therefore we may conclude that there is no subset $A\subset \mathbb{R}$ such that $A'=\mathbb{Q}$.

$\Box$

Reflection: This was a really nice question, especially because on the actual qual they gave us the definition of a cluster point. The main thing to take away from this problem is that once you have the rationals in your set of cluster points, you really have to have all of $\mathbb{R}$ since the rationals are dense in $\mathbb{R}$.