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 $\text{int}\mathbb{(Q)} = \mathbb{Q}$?
 A: 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.
A: I'm assuming since you're using the Euclidean Metric that you're viewing $\mathbb{Q}$ as a subset of $\mathbb{R}$.  The emptiness of the interior follows from the density of the rationals in the reals.  So in fact, you can't actually take an open set around a rational number and stay within the rationals because real numbers will always get in your way.
A: The interior of rational numbers is not empty when the whole set is also rational numbers.
But if the whole set is real numbers, then any rational numbers would have a ball containing real numbers. By definition, this rational number is not an interior point because its ball contains real numbers which are not part of the set of rational numbers.
A: Think of it this way: we say $A$ is a subset of $B$ if for all $a \in A$, $a$ is in $B$.
Now, we say that the interior of some set $S$ is the set of all of its interior points. A point is an interior point if there exists a neighborhood $N$ of $x$, for some $x \in S$, such that $N$ is a subset of $S$.
Now, keeping in mind the definition of a subset, there exists no $\epsilon$ such that an irrational number is not contained within the neighborhood $(x - \epsilon, x + \epsilon)$ - hence, since there exists an $x \in N$ such that $x$ is not in $S$, given $I = \mathbb R \setminus \mathbb Q$, then there can be no such $N$ such that $N$ is a subset of $S$ - it fails to meet the definition of a subset. 
When you consider what is required for a point to be considered an interior point, you can see that no such interior points exist, and thus $\operatorname{int}(\mathbb Q) = \emptyset$. 
A: *

*Definition:
An element x is the interior point of A(subset of X) if there exists
open set U containing x such that U contained in A.

*Let x=2, A=Q, X=R(Real Numbers),U=(1,3)  Apply them on definition.

*The element 2 is interior point of Q if the open set U=(1,3) and 2 belongs to U such that (1,3)contained in Q.

*Look at the condition (bold line).. Do we have (1,3) contained in Q ? No ,since (1,3) contains an irrational number root2(root 2).
but which doesn't belongs to Q.

*Hence i can find an open set containing 2 but which not satisfies the condition (Bold one). So int Q = empty.

*it's understood that it will not work for any Q's.

*Hence int Q= empty.

A: 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 $\textrm{int}(\mathbb{Q}) = \emptyset$.
