Interior Set of Rationals. Confused!

Question

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 $\text{int}\mathbb{(Q)} = \mathbb{Q}$?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $\operatorname{int}(\mathbb Q) = \emptyset$.

Answer 5

1. 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.
2. Let x=2, A=Q, X=R(Real Numbers),U=(1,3) Apply them on definition.
3. 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.
4. 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.
5. Hence i can find an open set containing 2 but which not satisfies the condition (Bold one). So int Q = empty.
6. it's understood that it will not work for any Q's.
7. Hence int Q= empty.

Answer 6

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