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 $\text{int}\mathbb{(Q)} = \mathbb{Q}$?
Best Answer
If the whole set is $\mathbb{Q}$, then $\text{int}(\mathbb{Q})=\mathbb{Q}$,
If the whole set is $\mathbb{R}$ or $\mathbb{R}^n$, then $\text{int}(\mathbb{Q})=\emptyset$,
because, $\forall q\in \mathbb{Q}, and \,\forall \epsilon>0, B_\epsilon(q)=\{x\in\mathbb{R}:|x-q|<\epsilon\}$ contains irrational numbers, which are not in the $\mathbb{Q}$, so $q$ is not a interior point of $\mathbb{Q}$.
the statement is proved.
the problem depends on the whole set you are talking about.