Interior Set of Rationals. Confused! Can someone explain to me why the interior of rationals is empty? That is $\text{int}(\mathbb{Q}) = \emptyset$?
The definition of an interior point is "A point $q$ is an interior point of $E$ if there exists a ball at $q$ such that the ball is contained in $E$" and the interior set is the collection of all interior points.
So if I were to take $q = \frac{1}{2}$, then clearly $q$ is an interior point of $\mathbb{Q}$, since I can draw a ball of radius $1$ and it would still be contained in $\mathbb{Q}$. 
And why can't I just take all the rationals to be the interior? 
So why can't I have $\text{int}\mathbb{(Q)} = \mathbb{Q}$?

It is easy to show that there are irrational numbers between any two rational numbers. Let $q_1 < q_2$ be rational numbers and choose a positive integer $m$ such that $m(q_2-q_1)>2$. Taking some positive integer $m$ such that $m(b-a)>2$, the irrational number $m q_1+\sqrt{2}$ belongs to the interval $(mq_1, mq_2)$ and so the irrational number $q_1 + \frac{\sqrt{2}}{m}$ belongs to $(q_1,q_2)$.
With this in mind, there are irrational numbers in any neighbourhood of a rational number, which implies that $\textrm{int}(\mathbb{Q}) = \emptyset$.
