Why the interior of $\mathbb{Q}$ in $\mathbb{R}$ is empty? I don't understand why the interior of $\mathbb{Q}$ in $\mathbb{R}$ is empty, since, for every ball with the center being a rational number, given an $\epsilon>0$, I can find an infinite sequence of rational numbers that approach this point. For example, take $\frac{1}{2}$. The sequence $\frac{1}{2}+\frac{1}{n}$ can be made as close as I want to the number $\frac{1}{2}$, therefore I can always have open balls with center $\frac{1}{2}$ such that there are rationals inside it.
 A: It doesn’t matter that there are rationals inside the ball: what matters is that your open ball is not a subset of $\Bbb Q$. In order for $\Bbb Q$ to be open in $\Bbb R$, for each $q\in\Bbb Q$ there would have to be an open ball $B(q,\epsilon)$ about $q$ such that $B(q,\epsilon)\subseteq\Bbb Q$, and that is never the case: every open interval in $\Bbb R$ contains irrational numbers.
A: For $x$ to be an interior point, it is not enough that there exist some rational in the ball around $x$, the definition of interior point require that there is some ball that every points in it are rational.
Hint: Can you see that your proposed construction of the ball always contain some irrational points?
A: Let $x\in\mathbb{Q}$ Consider any open ball $\mathbb{B}(x,r)$ irrespective of how small you choose the radius $r$ the open ball $\mathbb{B}(x,r)\not\subset\mathbb{Q}$.Because between any two rations there are infinitely many irrations. That is why $int\mathbb{Q}=\phi$ in$\mathbb{R}$.
A: It is an immediate consequence of Baire's category theorem. Every singleton $\{q\}$ with $q$ rational number is a closed set with empty interior, therefore also the union of all these singletons, i.e., $\mathbb{Q}$, has empty interior.
