This is a question from Analysis by Beals: show there exists a proper subset $A$ of $\mathbb{Q}$ with empty boundary (my definition of boundary is $\bar{A} \cap \bar{A^c}$, ie. intersection of the closure with the closure of the complement).
I'm having some trouble thinking about this. I was thinking about taking a singleton $a$, which is its own closure, but then we could construct a sequence $a - 1/n$ which converges to $a$, so $a$ would be in the closure of its complement and thus would not have empty boundary. I tried a few other things but cannot get to an answer. Any suggestions would be welcome.