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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 = 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 $\text{int}\mathbb{Q} = \mathbb{Q}$

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Hint: what does contained in mean? –  Alex Olssen Oct 30 '12 at 4:42
    
notice that "the ball is contained in $E$" means all of the points in the ball is contained in $E$, but any "ball" centered at a rational point contain some irrational point, which is not in $\mathbb{Q}$ –  Tianyu Oct 30 '12 at 4:44
    
@Tao, but why do we even need to check those irrationals?Our ball will have holes, but we can't have that? –  sidht Oct 30 '12 at 4:47
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I'm assuming you are using the standard definition, a "ball" on the real line, centered at a point $x$ with radius $r$ is defined (as long as I know)to be the set $\{y: |y-x|<r\}$, certainly in this definition we cannot have "holes" –  Tianyu Oct 30 '12 at 4:53
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2 Answers

up vote 5 down vote accepted

If the whole set is the $\mathbb{Q}$, then $int\mathbb{Q}=\mathbb{Q}$,

If the whole set is the $\mathbb{R}$ or $\mathbb{R}^n$, then $int\mathbb{Q}=\emptyset$,

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.

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Oh okay, that makes more sense! –  sidht Oct 30 '12 at 5:36
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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.

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