Call a set $D\subseteq \mathbb{R}$ discrete if for each $x\in D$ there is $\epsilon>0$ such that $(-\epsilon,\epsilon)\cap D=\{x\}$.
Question: Is there a discrete subset $D\subseteq [0,1]$ such that its closure has positive Lebesgue measure?
It's known that the closure of a discrete set is nowhere dense (under suitable prerequisites): In a metric space with no isolated points, why is the closure of a discrete set nowhere dense?
However, there are closed nowhere dense sets of positive measure (e.g. fat Cantor sets). But I don't know if these are closures of discrete sets.
Best Answer
Let $C\subset {\mathbb R}$ be a Cantor set; let ${\mathcal J}$ denote the set of complementary intervals of $C$. In each complementary interval $I\in {\mathcal J}$ pick one point, $x_I\in I$. Lastly, take $X=\{x_I: I\in {\mathcal J}\}$. I will leave the rest to you to figure out.